Answer:
C₂ = 2.22 KJ/kg °C
Explanation:
Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:
[tex]Heat\ Lost\ by\ Metal\ Block = Heat\ Gained\ by\ Ice\\m_{1}C_{1} \Delta T_{1} = m_{2}C_{2} \Delta T_{2}[/tex]
where,
m₁ = mass of ice = 1 kg
C₁ = specific heat of ice = 2.04 KJ/kg.°C
ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C
m₂ = mass of metal block = 1 kg
C₂ = specific heat of metal = ?
ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C
Therefore, using these values in the equation, we get:
[tex](1\ kg)(2.04\ KJ/kg.^0C)(15.12\ ^0C) = (1\ kg)C_{2}(13.88\ ^0C) \\C_{2} = \frac{30.84\ KJ}{13.88\ kg/^0C}[/tex]
C₂ = 2.22 KJ/kg °C