Home » Question Bank » Measurement of temperature and heat
Physics

Measurement of temperature and heat

Chapter: Heat and Its Transmission

Question 1 of 5 NDA MCQ

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?

Detailed Explanation

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 5 NDA MCQ

The temperature of a body increases from 310 K to 340 K. The temperature increase in degree Celsius is

Detailed Explanation

Correct Answer

✓ Option B is correct. The temperature increase is 30°C.

📖 Core Concept

The Celsius and Kelvin scales are related as they share the same incremental temperature change, where a change of 1 K equals a change of 1°C.

🧪 Step-by-Step Breakdown

Given initial and final temperatures:

\[ T_1 = 310\,K, \quad T_2 = 340\,K \]

Calculate the temperature increase in Celsius:

\[ \Delta T_{C} = T_2 - T_1 \]
\[ \Delta T_{C} = 340\,K - 310\,K \]
\[ \Delta T_{C} = 30\,^{\circ}C \]

Thus, the temperature increase is 30°C.

🔍 Option Analysis

  • Option A: Incorrect, it suggests a 20°C increase.
  • Option C: Incorrect, it suggests a 37°C increase.
  • Option D: Incorrect, it suggests a 67°C increase.

⚡ Mnemonic / Speed-Run

Remember: To convert temperature change, use \(\Delta T(K) = \Delta T(°C)\); increments are equal.

6. Visual Suggestion

6. Visual Suggestion

A schematic diagram illustrating the Kelvin and Celsius scales with a zero-offset would help visualize the direct relationship between temperature increments.

📖 Factual Verification & Reference:

Verified according to NCERT Class XI Physics, Chapter 1 (Units and Measurement).

Question 3 of 5 NDA MCQ

Which one of the following instruments can be used to measure –250°C temperature?

Detailed Explanation

Correct Answer

✓ Option (d) is correct. Thermocouple-based thermometers can accurately measure down to -250°C.

📖 Governing Law / Core Concept

Thermocouples operate based on the Seebeck effect, which describes how a voltage is generated when two different metals are joined at two junctions that are at different temperatures.

🧪 Step-by-Step Breakdown

Other common temperature measuring devices have limitations:

  • Mercury Thermometer: Useful only down to approximately -38°C.
  • Alcohol Thermometer: Effective down to around -114°C, thus unsuitable for -250°C.
  • Clinical Thermometer: Measures between 32°C and 42°C; not applicable for such low temperatures.

Thermocouples, however, can function within a wide temperature range, from about -270°C (near absolute zero) to several thousand degrees Celsius.

🔍 Option Analysis

Each alternative thermometer mentioned has a maximum and minimum limit, which disqualifies them for measuring extremely low temperatures like -250°C. Specifically:

  • A: Mercury thermometers freeze above -38°C.
  • B: Alcohol thermometers freeze around -114°C.
  • C: Clinical thermometers do not cover low temperature ranges.
  • D: Thermocouples cater to extremely low temperatures.

⚡ Mnemonic / Speed-Run

Memory Aid: "Mercury melts, Alcohol is chill, Clinical is warm, Thermocouple thrills!"

6. Visual Suggestion

6. Visual Suggestion

A schematic diagram illustrating a thermocouple setup, showing two different metal wires joined at junctions, with temperature measurement zones marked. Emphasize the wide measurement range extending from -270°C to higher temperatures.

📖 Factual Verification & Reference

📖 Factual Verification & Reference:

Verified against NCERT Class XII Physics, Chapter on Thermodynamics and Temperature Measurement Instruments.

Question 4 of 5 NDA MCQ

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

Detailed Explanation

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 5 of 5 NDA MCQ

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

Detailed Explanation

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.

Unlock 10,000+ Practice Questions

This is a preview of the SenaPrep Question Bank. To access comprehensive chapter-wise exercises, bookmark questions, attempt custom tests, and track your progress, download the official SenaPrep Android App.

Download Android App