Question

A 100 gram glass container contains 250 grams of water at 15.0 °C. A 200 gram piece of lead at 100 °C is added to the water in the container. What is the final temperature of the system? (specific heat of water = 4,186 J/kg °C, specific heat of glass = 837.2 J/kg °C, specific heat of lead = 127.7 J/kg °C)

Answer 1

Let the temperature be T.

Heat lost by lead will be gained by the glass and water.

So,

Heat gain by Glass + Heat gain by water = Heat lost by lead

then equation will be,

$\frac{100}{1000}\times 837.2(T - 15) + \frac{250}{1000}\times 4186(T - 15) = \frac{200}{1000}\times 127.7(100 - T)$

Solving it, we get, $T = 16.88^{\circ} C$ which is the final solution of the system.