Question

2 kg of a substance receives 500 kJ heat and undergoes a temperature change from 100 $^\circ$C to 200 $^\circ$C. The average specific heat of the substance during the process (in kJ/kg K) will be

• A. \( 5 \)
• B. \( 10 \)
• C. \( 25 \)
• D. \( 2.5 \)

Hint:

The fundamental formula for heat transfer related to specific heat is \( Q = mc\Delta T \). Ensure consistent units, especially for temperature change (either \(^\circ\)C or K for \( \Delta T \) works as the magnitude of change is the same).

Answer 1

The Correct Option is D

Step 1: Identify the given values.
Mass of the substance \( m = 2 \text{ kg} \).
Heat received \( Q = 500 \text{ kJ} \).
Initial temperature \( T_1 = 100 \text{ \(^\circ\)C} \).
Final temperature \( T_2 = 200 \text{ \(^\circ\)C} \).
Step 2: Calculate the change in temperature.
The temperature change \( \Delta T = T_2 - T_1 \).
\[ \Delta T = 200 \text{ \(^\circ\)C} - 100 \text{ \(^\circ\)C} = 100 \text{ \(^\circ\)C} \] Note that a change in temperature of 100 \(^\circ\)C is equivalent to a change of 100 K. So, \( \Delta T = 100 \text{ K} \).
Step 3: Use the formula for heat transfer to find the specific heat.
The amount of heat transferred \( Q \) to a substance of mass \( m \) undergoing a temperature change \( \Delta T \) is given by: \[ Q = m c \Delta T \] where \( c \) is the average specific heat of the substance. Rearrange the formula to solve for \( c \): \[ c = \frac{Q}{m \Delta T} \]
Step 4: Substitute the values and calculate the specific heat. \[ c = \frac{500 \text{ kJ}}{2 \text{ kg} \times 100 \text{ K}} \] \[ c = \frac{500}{200} \text{ kJ/kg K} \] \[ c = 2.5 \text{ kJ/kg K} \] The final answer is $\boxed{\text{4}}$.