Home » Question Bank » Measurement of temperature and heat
Physics

Measurement of temperature and heat Questions

Chapter: Heat and Its Transmission | Total Available: 22 Real Questions

Question 1 of 10 Authentic UPSC NDA Question

What is the mass of a material, whose specificheat capacity is 400 J/(kg°C) for a rise intemperature from 15°C to 25°C, when heatreceived is 20 kJ?

Step-by-Step Logic & Solution

Solution

Option (d) is correct.

Explanation:

\[ q = mc\Delta T \]
In this formula, \( q \) is the heat absorbed in joules (J), \( m \) is the mass in kilograms (kg), \( c \) is the specific heat capacity in J/(kg°C), and \( \Delta T \) is the change in temperature in °C. Given: - \( q = 20,000 \, \text{J} \) (since 20 kJ = 20,000 J) - \( c = 400 \, \text{J/(kg°C)} \) - \( \Delta T = 25\,°C - 15\,°C = 10\,°C \) Rearranging the formula to calculate mass \( m \):
\[ m = \frac{q}{c \Delta T} \]
Substituting in the given values:
\[ m = \frac{20,000 \, \text{J}}{400 \, \text{J/(kg°C)} \cdot 10\,°C} \]
Calculating:
\[ m = \frac{20,000}{4000} = 5 \, \text{kg} \]
📌 Hints / Properties Used:
  • Specific heat capacity formula: \( q = mc\Delta T \)
  • Convert kJ to J when necessary: \( 1 \, \text{kJ} = 1000 \, \text{J} \)
📊 Visual Diagram Suggestion:

A diagram could illustrate the relationship between heat, mass, and temperature change: with a flow from heat energy (in J) through specific heat capacity (in J/(kg°C)) to the calculation of mass (in kg) alongside a thermometer indicating temperatures of 15°C and 25°C.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Thermodynamics.

Question 2 of 10 Authentic UPSC NDA Question

The temperature at which “vapour pressure of the liquid in an open vessel becomes equal tothe atmospheric pressure” is called

Step-by-Step Logic & Solution

Correct Answer

✓ Option B is correct. The boiling point is the temperature at which the vapor pressure of a liquid equals atmospheric pressure.

📖 Governing Law / Core Concept

The temperature at which the vapor pressure of a liquid equals the atmospheric pressure is defined as the boiling point.

🧪 Step-by-Step Breakdown

At boiling point:

\[ P_{vapor} = P_{atmospheric} \]

Where:

  • Pvapor = Vapor pressure of the liquid (measured in pascals, Pa)
  • Patmospheric = Atmospheric pressure (assumed standard value = 101.3 kPa)

When \( P_{vapor} \) reaches \( P_{atmospheric} \), bubbles of vapor form within the liquid, resulting in boiling.

🔍 Option Analysis

  • A: Melting point - The temperature at which a solid becomes a liquid.
  • C: Liquid point - Not a standard term in thermodynamics.
  • D: None of the above - This is incorrect as option B accurately identifies the boiling point.

⚡ Mnemonic / Speed-Run

To remember boiling point concepts: "Bubbles Boil at Boiling Point".

6. Visual Suggestion

6. Visual Suggestion

A diagram illustrating a boiling liquid in a pot, showing temperature arrows, vapor pressure, and atmospheric pressure lines, helping visualize how boiling occurs when vapor pressure equals atmospheric pressure.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class 11 Physics, Chapter 1 (Thermodynamics).

Question 3 of 10 Authentic UPSC NDA Question

Numerically two thermometers, one in Fahrenheit scale and another in Celsius scale shall read same at

Step-by-Step Logic & Solution

Correct Answer

✓ Option (a) is correct. Both scales read -40° at the same temperature.

📖 Governing Law / Core Concept

The relationship between Fahrenheit (\( F \)) and Celsius (\( C \)) is given by the equation:
\[ F = \frac{9}{5}C + 32 \]

🧪 Step-by-Step Breakdown

Set \( F = C \):
\[ C = \frac{9}{5}C + 32 \]
Rearranging gives:
\[ C - \frac{9}{5}C = 32 \]
Simplification leads to:
\[ -\frac{4}{5}C = 32 \]
Thus,
\[ C = -40 \]

🔍 Option Analysis

- **B: 0** - Corresponds to freezing point, not equal. - **C: -273** - Absolute zero, not on the Fahrenheit scale. - **D: 100** - Boiling point, not equal.

⚡ Mnemonic / Speed-Run

Remember: \( C \) and \( F \) intersect at -40 for a quick reference.

6. Visual Suggestion

6. Visual Suggestion

A graph showing the linear relationship between Celsius and Fahrenheit scales, highlighting the intersection at (-40, -40).

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter on Thermodynamics.

Question 4 of 10 Authentic UPSC NDA Question

Which one of the following statements regarding a thermos flask is NOT correct ?

Step-by-Step Logic & Solution

Correct Answer

✓ Option (c) is correct. The inner wall does not radiate significant heat in a thermos.

📖 Core Concept

A thermos flask is designed to minimize heat transfer. It typically consists of two glass walls separated by a vacuum. This design significantly inhibits conduction and convection.

🧪 Step-by-Step Breakdown

The thermos flask prevents heat transfer primarily through:

  • Vacuum Insulation: Eliminates conduction and convection between walls.
  • Shiny Surface: The inner wall reflects radiant heat, minimizing loss.
  • Cork Insulation: Used to seal and provide additional thermal resistance.

Specifically, regarding Option (c):

The inner wall actually reflects rather than radiates heat due to its shiny surface, making this statement incorrect.

🔍 Option Analysis

  • Option (a): Correctly states the use of vacuum and glass, both poor heat conductors.
  • Option (b): The shiny surfaces of the glass walls are accurate for energy reflection.
  • Option (d): Cork supports serve as good insulators, thus this statement is true.

⚡ Mnemonic / Key Properties

Remember: "Vacuum is King!" for heat insulation in thermoses.

📊 Visual Suggestion

Visual Suggestion

A diagram illustrating the structure of a thermos flask, highlighting the vacuum layer, the shiny internal surface, and arrows indicating minimal heat loss via radiation, conduction, and convection.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XI Physics, Chapter 11 (Thermal Properties of Matter).

Question 5 of 10 Authentic UPSC NDA Question

In which of the following phenomena do heat waves travel along a straight line with the speed of light?

Step-by-Step Logic & Solution

Correct Answer

✓ Option (c) is correct. Heat waves travel through thermal radiation.

📖 Governing Law / Core Concept

Thermal radiation can be described by Stefan-Boltzmann Law:

\[ E = \sigma T^4 \]

Where \( E \) is the radiant energy, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the absolute temperature in Kelvin.

🧪 Step-by-Step Breakdown

In thermal conduction and convection, heat transfer requires a medium.

  • Thermal conduction: Transfer of heat through material without movement of material.
  • Thermal convection: Transfer of heat through the movement of fluids.
  • Thermal radiation: Transfer of energy via electromagnetic waves through a vacuum.

Radiation propagates at the speed of light, allowing it to travel through space.

🔍 Option Analysis

  • A: Thermal conduction requires a medium and cannot propagate through a vacuum.
  • B: Thermal convection relies on fluid motion; thus, it cannot occur in a vacuum.
  • D: Since both conduction and convection are incorrect, this option is false.

⚡ Mnemonic / Speed-Run

Remember "C for Conduction" and "V for Vacuum" - only Radiation can travel in a Vacuum.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 12 (Thermal Properties of Matter)

Question 6 of 10 Authentic UPSC NDA Question

10 g of ice at -10 °C is is mixed with 10 g of water at 0 °C. The amount of heat required to raise the temperature of mixture to 10 °C is

Step-by-Step Logic & Solution

Solution

Option (c) is correct.

Explanation:

\[ Q = m \cdot c \cdot \Delta T \]
\[ Q_{ice} = m_{ice} \cdot c_{ice} \cdot (0 - (-10)) + m_{ice} \cdot L_f + m_{water} \cdot c_{water} \cdot (10 - 0) \]
\[ Q_{ice} = 10 \cdot 0.5 \cdot (10) + 10 \cdot 80 + 10 \cdot 1 \cdot (10) \]
\[ = 50 + 800 + 100 \]
\[ = 950 \text{ cal} \]
\[ Q_{total} = Q_{ice} + Q_{water} \]
\[ Q_{total} = 950 + 100 = 1050 \text{ cal} \]
📌 Hints / Properties Used:
  • Specific heat of ice: \(c_{ice} = 0.5 \text{ cal/g°C}\)
  • Latent heat of fusion: \(L_f = 80 \text{ cal/g}\)
  • Specific heat of water: \(c_{water} = 1 \text{ cal/g°C}\)
📊 Visual Diagram Suggestion:

A schematic showing the phase change of ice to water, including temperature changes, highlighting energy transfer at each step from ice at -10°C to ice at 0°C, water heating from 0°C to 10°C, and heat values involved.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 11 (Thermodynamics).

Question 7 of 10 Authentic UPSC NDA Question

Water is heated with a coil of resistance R connected to domestic supply. The rise of temperature of water will depend on

(1) Supply voltage.

(2) Current passing through the coil.

(3) Time for which voltage is supplied. Select the correct answer from among the following : 

Step-by-Step Logic & Solution

Correct Answer

✓ Option (a) is correct. The temperature rise depends on all three factors: supply voltage, current through the coil, and time.

📖 Governing Law / Core Concept

The temperature rise in water due to the heating coil can be described by the formula:

\[ Q = V \cdot I \cdot t \text{, where } Q \text{ is heat energy (in J), } V \text{ is voltage (in V), } I \text{ is current (in A), and } t \text{ is time (in s).} \]

The heat energy transferred to the water leads to a temperature increase.

🧪 Step-by-Step Breakdown

The total heat energy supplied is given by:

\[ Q = m \cdot c \cdot \Delta T \text{ (where } m \text{ is mass of water, } c \text{ is specific heat capacity, and } \Delta T \text{ is temperature rise).} \]

From Ohm's Law, we have:

\[ V = I \cdot R \]

Substituting for \(I\) gives:

\[ Q = (V/R) \cdot V \cdot t = \frac{V^2 \cdot t}{R} \]

This shows temperature increase depends on voltage \(V\), current \(I\), and time \(t\).

🔍 Option Analysis

Option (b): Incorrect, only mentions voltage and current but omits time.

Option (c): Incorrect, only addresses voltage.

Option (d): Incorrect, combines only current and time but ignores voltage.

⚡ Mnemonic / Speed-Run

Remember: "VIT" = Voltage, Current, Time - all are crucial for heating water!

📊 Visual Suggestion

📊 Visual Suggestion

A flowchart displaying the relationship between heat energy (Q), voltage (V), current (I), resistance (R), and time (t) in heating processes. Include an illustration of a heating coil in a container of water to visually represent energy transfer.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter 12 (Thermodynamics).

Question 8 of 10 Authentic UPSC NDA Question

The temperature of a place on one sunny day is 113 in Fahrenheit scale. The Kelvin scale reading of this temperature will be

Step-by-Step Logic & Solution

Solution

Option (a) is correct.

Explanation:

The conversion from Fahrenheit (°F) to Kelvin (K) is given by the formula:
\[ K = \frac{5}{9}(°F - 32) + 273.15 \]
\[ K = \frac{5}{9}(113 - 32) + 273.15 \]
\[ K = \frac{5}{9}(81) + 273.15 \]
\[ K = 45 + 273.15 \]
\[ K \approx 318.15 \approx 318 \text{ K} \]
📌 Hints / Properties Used:
  • Fahrenheit to Kelvin conversion formula
  • Basic arithmetic operations
  • Understanding temperature scales
📊 Visual Diagram Suggestion:

A flowchart illustrating temperature conversion processes, showing steps to convert Celsius, Fahrenheit, and Kelvin, emphasizing the use of formulas at each point.

📖 Factual Verification & Reference:

Verified against NCERT Class IX Science, Chapter on Heat and Temperature.

Question 9 of 10 Authentic UPSC NDA Question

Radiations coming from the Sun are mostly in the form of

Step-by-Step Logic & Solution

Correct Answer

✓ Option (c) is correct. The Sun emits both visible light and short-wavelength infrared radiation.

📖 Governing Law / Core Concept

The electromagnetic spectrum describes the range of all possible frequencies of electromagnetic radiation, where visible light and infrared occupy a significant portion.

🧪 Step-by-Step Breakdown

Radiation from the Sun primarily consists of:

  • Visible Light: The light we can see.
  • Short-Wavelength Infrared: This includes infrared radiation just beyond the visible spectrum.

Most solar radiation reaches Earth in the form of:

\[ \text{Total Solar Radiation} \approx \text{Visible Light} + \text{Short-Wavelength Infrared} \]

🔍 Option Analysis

  • Option (a): Incorrect, as it excludes infrared.
  • Option (b): Incorrect, as it mentions long-wavelength infrared, which is not predominant.
  • Option (d): Incorrect, as it includes both types of infrared; only short-wavelength is primarily emitted.

⚡ Mnemonic / Speed-Run

Remember: "Visible shines, infrared aligns."

6. Visual Suggestion

6. Visual Suggestion

A diagram illustrating the electromagnetic spectrum, highlighting the visible light and short-wavelength infrared sections, would aid understanding.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Electromagnetic Waves.

Question 10 of 10 Authentic UPSC NDA Question
What is the primary mode of heat transfer in a vacuum?
Step-by-Step Logic & Solution
[CORE CONCEPT]: Heat transfer occurs through conduction, convection, and radiation. In a vacuum, conduction and convection cannot occur. [STEP-BY-STEP SOLUTION]: Radiation is the only mode of heat transfer that can occur in a vacuum, as it involves the transfer of energy through electromagnetic waves. [SHORTCUT / EXAM TRICK]: Remember that radiation is how the Sun's heat reaches Earth through the vacuum of space. [COMMON MISTAKE TRAP]: Students might confuse radiation with conduction or convection, thinking that heat can travel through empty space by other means.
Today's UPSC Current Affairs

Padma Bhushan Call Deepens India Election Commission Row - streamlinefeed.co.ke

Read Today's Digest