A cast iron weight weighing 100 g, heated to 100 degrees Celsius, was lowered into water weighing 400 g

A cast iron weight weighing 100 g, heated to 100 degrees Celsius, was lowered into water weighing 400 g, as a result of heat exchange, the weight cooled to 40 degrees Celsius. How many degrees did the water heat up?

Given:

m1 = 100 grams = 0.1 kilograms – the mass of the cast iron weight;

m2 = 400 grams = 0.4 kilograms – the mass of water;

T1 = 100 degrees Celsius – the initial temperature of the cast iron;

T2 = 40 degrees Celsius – the temperature of the cast iron weight after it was lowered into the water;

c1 = 540 J / (kg * C) – specific heat capacity of cast iron;

c2 = 4200 540 J / (kg * C) – specific heat capacity of water.

It is required to determine dT (degree Celsius) – how many degrees the water has heated up.

Let’s find the amount of heat that was released when cooling the cast-iron weight:

Q = c1 * m1 * (T1 – T2) = 540 * 0.1 * (100 – 40) = 54 * 60 = 3240 Joules.

This energy was used to heat water, that is:

Q = c2 * m2 * dT, hence:

dT = Q / (c2 * m2) = 3240 / (4200 * 0.4) = 3240/1680 = 1.9 degrees Celsius.

Answer: the water warmed up by 1.9 degrees Celsius.