1. Problem 1: Calculate the specific heat capacity of a metal block heated by an electric coil. 2. Given: - Power supplied, $P = 50$ W - Mass of metal block, $m = 0.60$ kg - Time, $t = 90$ s - Initial temperature, $T_i = 20^\circ$C - Final temperature, $T_f = 45^\circ$C 3. Formula used: The heat energy supplied is $Q = P \times t$. The heat absorbed by the metal is $Q = m c \Delta T$, where $c$ is the specific heat capacity and $\Delta T = T_f - T_i$. 4. Calculate heat energy supplied: $$Q = 50 \times 90 = 4500 \text{ J}$$ 5. Calculate temperature change: $$\Delta T = 45 - 20 = 25^\circ C$$ 6. Rearrange formula to find $c$: $$c = \frac{Q}{m \Delta T} = \frac{4500}{0.60 \times 25} = \frac{4500}{15} = 300 \text{ J/(kg K)}$$ 7. Assumption: No heat loss to the surroundings; all electrical energy converts to heat absorbed by the metal. --- 8. Problem 2: Find (a)(i) energy supplied by heater, (a)(ii) energy absorbed by calorimeter, (a)(iii) specific heat capacity of liquid, and (b) assumptions. 9. Given: - Mass of calorimeter, $m_c = 270$ g = 0.270 kg - Mass of liquid, $m_l = 260$ g = 0.260 kg - Time, $t = 360$ s - Initial temperature, $T_i = 18^\circ$C - Final temperature, $T_f = 30^\circ$C - Potential difference, $V = 12.0$ V - Current, $I = 3.4$ A - Specific heat capacity of copper, $c_c = 400$ J/(kg K) 10. (a)(i) Energy supplied by heater: $$Q = V I t = 12.0 \times 3.4 \times 360 = 14688 \text{ J}$$ 11. (a)(ii) Energy absorbed by calorimeter: Temperature change: $$\Delta T = 30 - 18 = 12^\circ C$$ Energy absorbed: $$Q_c = m_c c_c \Delta T = 0.270 \times 400 \times 12 = 1296 \text{ J}$$ 12. (a)(iii) Energy absorbed by liquid: $$Q_l = Q - Q_c = 14688 - 1296 = 13392 \text{ J}$$ Calculate specific heat capacity of liquid $c_l$: $$c_l = \frac{Q_l}{m_l \Delta T} = \frac{13392}{0.260 \times 12} = \frac{13392}{3.12} \approx 4292.31 \text{ J/(kg K)}$$ 13. (b) Assumptions: - No heat loss to surroundings. - The calorimeter and liquid reach thermal equilibrium. - Electrical energy fully converts to heat.