Answer:
[tex] 21.14 [/tex] °C
Explanation:
[tex]m_{i}[/tex] = mass of ice cube = 0.0600 kg
[tex]T_{i}[/tex] = initial temperature of ice cube = - 30.0°C
[tex]c_{i}[/tex] = specific heat of ice = 2030 Jkg⁻¹ C⁻¹
[tex]L[/tex] = Latent heat of fusion of ice to water = 334000 J/kg
[tex]T_{w}[/tex] = initial temperature of water = 35.0°C
[tex]c_{i}[/tex] = specific heat of water = 4186 Jkg⁻¹ C⁻¹
[tex]T_{f}[/tex] = Final equilibrium temperature = ?
Using conservation of heat
[tex]m_{i}c_{i}(0 - T_{i}) + m_{i} L + m_{i} c_{w}(T_{f} - 0) = m_{w} c_{w} (T_{w} - T_{f})[/tex]
[tex](0.060)(2030)(0 - (- 30)) + (0.060) (334000) + (0.060) (4186)(T_{f} - 0) = (0.500) (4186) (35 - T_{f})[/tex]
[tex]T_{f} = 21.14 [/tex] °C
