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:
The value is [tex]T_c = 12 .1 ^oC[/tex]
Explanation:
From the question we are told that
The mass of the ice cube is [tex]m_i = 7.50 *10^{-3} \ kg[/tex]
The temperature of the ice cube is [tex]T_i = 0^o C [/tex]
The mass of the copper cube is [tex]m_c = 0.540 \ kg[/tex]
The final temperature of both substance is [tex]T_f = 0^oC[/tex]
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
[tex]Q = m_c * c_c * [T_c – T_f ][/tex]
The specific heat of copper is [tex]c_c = 385J/kg \cdot ^oC[/tex]
Generally the heat gained by the ice cube is mathematically represented as
[tex]Q_1 = m_i * L [/tex]
Here L is the latent heat of fusion of the ice with value [tex]L = 3.34 * 10^{5} J/kg[/tex]
So
[tex]Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5} [/tex]
=> [tex]Q_1 = 2505 \ J [/tex]
So
[tex]2505 = 0.540 * 385 * [T_c – 0 ][/tex]
=> [tex]T_c = 12 .1 ^oC[/tex]