Question

A 0.0600 kg ice cube at -30.0°C is placed in 0.500 kg of 35.0°C water in a very well insulated container. What is the final temperature? °C

Answer 1

`21.14` °C

Step-by-step explanation:

`m_(i)` = mass of ice cube = 0.0600 kg

`T_(i)` = initial temperature of ice cube = - 30.0°C

`c_(i)` = specific heat of ice = 2030 Jkg⁻¹ C⁻¹

`L` = Latent heat of fusion of ice to water = 334000 J/kg

`T_(w)` = initial temperature of water = 35.0°C

`c_(i)` = specific heat of water = 4186 Jkg⁻¹ C⁻¹

`T_(f)` = Final equilibrium temperature = ?

Using conservation of heat

`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))`

`(0.060)(2030)(0 - (- 30)) + (0.060) (334000) + (0.060) (4186)(T_(f) - 0) = (0.500) (4186) (35 - T_(f))`

`T_(f) = 21.14` °C