A silver spoon weighing 30 g, having a room temperature of 25 degrees, was dropped into a glass containing

A silver spoon weighing 30 g, having a room temperature of 25 degrees, was dropped into a glass containing 200 g of water at a temperature of 90 degrees. What will the water temperature become after thermal equilibrium is established?

Given:

m1 = 200 grams = 0.2 kilograms is the mass of water;

T1 = 90 ° Celsius – water temperature;

m2 = 30 grams = 0.03 kilograms – the mass of a silver spoon;

T2 = 25 ° Celsius – temperature of the silver spoon;

c1 = 4200 J / (kg * C) – specific heat of water;

c2 = 250 J / (kg * C) – specific heat of silver.

It is required to determine T (degree Celsius) – the temperature of the water after thermal equilibrium is established.

After placing the silver spoon in the water, the water will cool and the spoon will heat up. Then:

Q1 = Q2;

c1 * m1 * (T1 – T) = c2 * m2 * (T – T2);

4200 * 0.2 * (90 – T) = 250 * 0.03 * (T – 25);

840 * (90 – T) = 7.5 * (T – 25);

75600 – 840 * T = 7.5 * T – 187.5;

75787.5 = 847.5 * T;

T = 75787.5 / 847.5 = 89.4 ° Celsius.

Answer: the water temperature will be equal to 89.4 ° Celsius.