Q139 / 150 · MHT CET 2026 (20 April Shift 2) · Physics
A piece of metal of 850 K is dropped into 1 kg of water at 300 K. If equilibrium temperature of water is 350 K then the heat capacity of the metal expressed in $\dfrac{J}{K}$ is $\left(\text{Sp. heat of water} = 1\ \dfrac{\text{cal}}{\text{g}^\circ\text{C}}\right)$
- Correct answerA$420$
- B$240$
- C$100$
- D$24$
Solution
According to the principle of calorimetry, heat lost by the metal is equal to the heat gained by the water.
Heat gained by water is $Q = m_w s_w \Delta T_w$
Given mass of water $m_w = 1\text{ kg} = 1000\text{ g}$
Specific heat of water $s_w = 1\text{ cal/g}^\circ\text{C} = 4.2\text{ J/g}^\circ\text{C}$
Change in temperature of water $\Delta T_w = 350 - 300 = 50\text{ K}$
$Q = 1000 \times 4.2 \times 50 = 210000\text{ J}$
Heat lost by the metal is $Q = C \Delta T_m$
Change in temperature of the metal $\Delta T_m = 850 - 350 = 500\text{ K}$
Equating the heat lost to the heat gained:
$C \times 500 = 210000$
$C = \dfrac{210000}{500} = 420\text{ J/K}$
Answer: $420$
Heat gained by water is $Q = m_w s_w \Delta T_w$
Given mass of water $m_w = 1\text{ kg} = 1000\text{ g}$
Specific heat of water $s_w = 1\text{ cal/g}^\circ\text{C} = 4.2\text{ J/g}^\circ\text{C}$
Change in temperature of water $\Delta T_w = 350 - 300 = 50\text{ K}$
$Q = 1000 \times 4.2 \times 50 = 210000\text{ J}$
Heat lost by the metal is $Q = C \Delta T_m$
Change in temperature of the metal $\Delta T_m = 850 - 350 = 500\text{ K}$
Equating the heat lost to the heat gained:
$C \times 500 = 210000$
$C = \dfrac{210000}{500} = 420\text{ J/K}$
Answer: $420$
Hence, the correct answer is option: 'A'.