Question

A solid 1.572 kg gold container with an initial temperature of 20.3 degrees C holds 0.674 kg of liquid ethyl alcohol at 35.8 degrees C. the specific heat of gold is 129 j/kgC and the specific heat of ethyl alcohol is 2,450 j/kgC. Assume the system is isolated. Calculate the final, equilibrium temperature of the system

Answer 1

Answer on Question #69597 Physics / Other

A solid $m_{1} = 1.572 \, \mathrm{kg}$ gold container with an initial temperature of $t_{1} = 20.3$ degrees C holds $m_{2} = 0.674 \, \mathrm{kg}$ of liquid ethyl alcohol at $t_{2} = 35.8$ degrees C. the specific heat of gold is

$c_{1} = 129 \, \mathrm{j/kgC}$ and the specific heat of ethyl alcohol is $c_{2} = 2.450 \, \mathrm{j/kgC}$. Assume the system is isolated. Calculate the final, equilibrium temperature of the system.

Solution:

Let us denote as $t$ the final temperature.

The energy balance equation

$$c_{1} m_{1} (t - t_{1}) = c_{2} m_{2} (t_{2} - t)$$

Thus

$$t = \frac{c_{1} m_{1} t_{1} + c_{2} m_{2} t_{2}}{c_{1} m_{1} + c_{2} m_{2}} = \frac{129 \times 1.572 \times 20.3 + 2450 \times 0.674 \times 35.8}{129 \times 1.572 + 2450 \times 0.674} = \frac{63233.1364}{1854.088} = 34.1{}^{\circ}\mathrm{C}$$

Answers: $t = 34.1{}^{\circ}\mathrm{C}$

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