Question

If 100 kg of water is to be cooled from $80^\circ$C to $30^\circ$C, then how much heat is removed? (Assumed specific heat of water is 1 kcal/kg$^\circ$C)

• A. 3900 kJ
• B. 21000 kJ
• C. 20900 kJ
• D. 4180 kJ

Hint:

Always remember: $Q = mc\Delta T$ and $1$ kcal $= 4.2$ kJ.

Answer 1

The Correct Option is B

Concept: The amount of heat removed during cooling of a substance is calculated using the formula: \[ Q = m \, c \, \Delta T \] where:
• $Q$ = heat removed
• $m$ = mass
• $c$ = specific heat capacity
• $\Delta T$ = temperature difference

Step 1: Writing given data clearly.
• Mass, $m = 100$ kg
• Initial temperature = $80^\circ$C
• Final temperature = $30^\circ$C
• Specific heat, $c = 1$ kcal/kg$^\circ$C

Step 2: Calculating temperature difference. \[ \Delta T = 80 - 30 = 50^\circ C \]

Step 3: Applying heat equation. \[ Q = 100 \times 1 \times 50 = 5000 \text{ kcal} \]

Step 4: Converting kcal to kJ.
We know: \[ 1 \text{ kcal} = 4.2 \text{ kJ} \] \[ Q = 5000 \times 4.2 = 21000 \text{ kJ} \]

Step 5: Matching with options.
Option (2) matches the calculated value. Final Conclusion:
Heat removed is 21000 kJ. Hence, the correct answer is option (2).