Question

A metal plate of mass 50 g is heated to a temperature of $200^\circ\mathrm{C}$ and then immersed in 150 g of water at $20^\circ\mathrm{C}$. If the temperature of the water is $50^\circ\mathrm{C}$ at equilibrium, then the specific heat capacity of the metal is (specific heat capacity of water $=4200\ \mathrm{J\,kg^{-1}K^{-1}}$):

• A. $2970\ \mathrm{J\,kg^{-1}K^{-1}}$
• B. $2520\ \mathrm{J\,kg^{-1}K^{-1}}$
• C. $5040\ \mathrm{J\,kg^{-1}K^{-1}}$
• D. $1260\ \mathrm{J\,kg^{-1}K^{-1}}$

Hint:

Always convert masses to kg and temperatures to K differences (°C differences are same as K differences) when using $Q=mc\Delta T$.
Balance heat lost = heat gained (isolated system assumption).
Check units at each step to avoid arithmetic mistakes.
Round only at final step to maintain accuracy.

Answer 1

The Correct Option is B

• Convert masses to kilograms: metal $m_m = 50\ \text{g} = 0.05\ \text{kg}$, water $m_w = 150\ \text{g} = 0.15\ \text{kg}$.
• Initial temperatures: $T_{m,i}=200^\circ\mathrm{C}$, $T_{w,i}=20^\circ\mathrm{C}$. Final equilibrium temperature: $T_f=50^\circ\mathrm{C}$.
• Heat lost by metal $= m_m c_m (T_{m,i}-T_f)$.
• Heat gained by water $= m_w c_w (T_f - T_{w,i})$ where $c_w = 4200\ \mathrm{J\,kg^{-1}K^{-1}}$.
• Equate heat lost and gained (neglecting losses to environment): \[ m_m c_m (200-50) = m_w c_w (50-20). \] • Substitute numbers: \[ 0.05\,c_m \times 150 = 0.15 \times 4200 \times 30. \] \[ 7.5\,c_m = 0.15 \times 4200 \times 30 = 18900. \] \[ c_m = \frac{18900}{7.5} = 2520\ \mathrm{J\,kg^{-1}K^{-1}}. \] • Hence the specific heat capacity of the metal is $2520\ \mathrm{J\,kg^{-1K^{-1}}$}.