A 5.80kg piece of solid copper metal at an initial temperature T is placed with 2.00 kg of ice that is initially at -17.0?C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 0.80kg of ice and 1.20kg of liquid water.What was the initial temperature of the piece of copper?Express your answer to three significant figures and include the appropriate units.

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klefenz klefenz · Answer 1 · 2019-09-08T23:24:10+02:00

Answer:

208 C

Explanation:

First we need the specific heat capacity of copper.

[tex]Cp(Cu) = \frac{0.092 kca}{kg * C}[/tex]

The specific heat capacity for ice is:

[tex]Cp(ice) = \frac{0.45 kca}{kg * C}[/tex]

And the latent heat of melting ice into water is:

[tex]Cl(ice) = \frac{80 kca}{kg}[/tex]

Since the container has water and ice together, they must be at 0 C (assuming normal atmospheric pressure), and the copper piece too.

Therefore, all the ice heated from -17C to 0C. For this it took an energy of

[tex]Q1 = m * Cp(ice) * (tfinal - ti) = 2 * 0.45 * (0 - (-17)) = 15.3 kcal[/tex]

Also, part of the ice melted into water, this consumed energy too:

[tex]Q2 = m2 * Cl(ice) = 1.2 * 80 = 96 kcal[/tex]

The copper piece provided that heat, it released it by cooling down.

[tex]-(Q1 + Q2) = m(Cu) * Cp(Cu) * (tfinal(Cu) - ti(Cu))[/tex]

The copper ended at 0C.

[tex]-(Q1 + Q2) = m(Cu) * Cp(Cu) * (-ti(Cu))[/tex]

[tex]Q1 + Q2 = m(Cu) * Cp(Cu) * ti(Cu)[/tex]

[tex]ti(Cu) = \frac{Q1 + Q2}{m(Cu) * Cp(Cu)}[/tex]

[tex]ti(Cu) = \frac{15.3 + 96}{5.8 * 0.092} = 208 C[/tex]