Question

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts, and the final equilibrium temperature of the two substances is 0.00∘C. What was the initial temperature of the copper cube? Assume no heat is exchanged with the surroundings.

Answer 1

The value is
`T_c = 12 .1 ^oC`

Step-by-step explanation:

From the question we are told that

The mass of the ice cube is
`m_i = 7.50 *10^(-3) \ kg`

The temperature of the ice cube is
`T_i = 0^o C`

The mass of the copper cube is
`m_c = 0.540 \ kg`

The final temperature of both substance is
`T_f = 0^oC`

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

`Q = m_c * c_c * [T_c - T_f ]`

The specific heat of copper is
`c_c = 385J/kg \cdot ^oC`

Generally the heat gained by the ice cube is mathematically represented as

`Q_1 = m_i * L`

Here L is the latent heat of fusion of the ice with value
`L = 3.34 * 10^(5) J/kg`

So

`Q_1 = 7.50 *10^(-3) * 3.34 * 10^(5)`

=>
`Q_1 = 2505 \ J`

So

`2505 = 0.540 * 385 * [T_c - 0 ]`

=>
`T_c  =  12 .1 ^oC`