Question

The heat energy released by water of mass 2 kg when it is cooled by $10~^\circ\text{C}$ is (Specific heat capacity of water $= 4200 \text{ J kg}^{-1} \text{K}^{-1}$)

• A. $42000 \text{ J}$
• B. $21000 \text{ J}$
• C. $63000 \text{ J}$
• D. $84000 \text{ J}$

Hint:

Tip: For temperature changes, $\Delta T$ in Kelvin or Celsius is numerically equal, so use directly in the formula.

Answer 1

The Correct Option is D

Use the heat formula: $Q = mc\Delta T$. Given: $m = 2 \text{ kg}, \Delta T = 10^\circ C = 10 \text{ K}, c = 4200 \text{ J/kg·K}$. Substituting: \[ Q = 2 \cdot 4200 \cdot 10 = 84000 \text{ J} \] This is the amount of heat lost when water cools down by $10^\circ C$.

Answer 2

Step 1: Identify given data
Mass of water, \(m = 2 \text{ kg}\)
Temperature change, \(\Delta T = 10^\circ \text{C}\)
Specific heat capacity of water, \(c = 4200 \text{ J kg}^{-1} \text{K}^{-1}\)

Step 2: Use the formula for heat energy released or absorbed
\[ Q = mc\Delta T \]

Step 3: Calculate heat energy released
\[ Q = 2 \times 4200 \times 10 = 84000 \text{ J} \]

Step 4: Conclusion
The heat energy released by 2 kg of water when cooled by \(10^\circ \text{C}\) is \(84000 \text{ J}\).