A 4-kg mass of metal of unknown specific heat at a temperature of 600 $\circ$ C is dropped into 0.5 kg of ice and 0.5 kg of water both at 0 $\circ$ C.With no heat losses to the surroundings,the equilibrium temperature of the mixture is 85 $\circ$ C.Calculate the specific heat of the metal.
A)0.04 kcal/kg $\circ$ C
B)0.06 kcal/kg $\circ$ C
C)0.08 kcal/kg $\circ$ C
D)1.6 kcal/kg $\circ$ C
E)1.2 kcal/kg $\circ$ C

Question

Quizplus · Accepted Answer

To find the specific heat of the metal, we use the principle of conservation of energy. The heat lost by the metal will be equal to the heat gained by the ice and water mixture to reach the equilibrium temperature. First, convert temperatures to Celsius for consistency:- Initial temperature of metal: $600^\circ F = \frac{5}{9}(600-32) = 315.56^\circ C$- Equilibrium temperature: $85^\circ F = \frac{5}{9}(85-32) = 29.44^\circ C$The heat required to melt 0.5 kg of ice to water at $0^\circ C$ is:$$Q_{melt} = m_{ice} \cdot L_f = 0.5 \cdot 80 = 40 \, kcal$$ (where $L_f = 80 \, kcal/kg$ is the latent heat of fusion of ice)The heat required to raise the temperature of 1 kg of water (0.5 kg from ice + 0.5 kg initially) from $0^\circ C$ to $29.44^\circ C$ is:$$Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T = 1 \cdot 1 \cdot 29.44 = 29.44 \, kcal$$ (where $c_{water} = 1 \, kcal/kg^\circ C$ is the specific heat capacity of water)The total heat gained by the water and melted ice is:$$Q_{total} = Q_{melt} + Q_{water} = 40 + 29.44 = 69.44 \, kcal$$This heat is lost by the metal, so:$$Q_{metal} = m_{metal} \cdot c_{metal} \cdot \Delta T_{metal} = 69.44 \, kcal$$Solving for $c_{metal}$:$$c_{metal} = \frac{69.44}{4 \cdot (315.56 - 29.44)} = \frac{69.44}{1144.48} = 0.0607 \, kcal/kg^\circ C$$Thus, the specific heat of the metal is approximately $0.06 \, kcal/kg^\circ C$, which corresponds to choice B.

A 4-Kg Mass of Metal of Unknown Specific Heat at a Temperature