A metal cylinder with a mass of 200 g was heated in boiling water to 100 * C and then lowered into

A metal cylinder with a mass of 200 g was heated in boiling water to 100 * C and then lowered into water with a mass of 400 g having a temperature of 22 * C. After a while the temperature of the water and the cylinder became equal to 25 * C. What is the specific heat of the metal from which the cylinder is made? neglected.

Since a cylinder heated to 100 degrees is placed in cold water, therefore, heat exchange occurs between the water and the cylinder. Therefore, the amount of heat transferred from the cylinder is equal to:

Q1 = c1 * m1 * (t1 – t).

Likewise for water:

Q2 = c2 * m2 * (t – t2).

c1 * m1 * (t1 – t) = c2 * m2 * (t – t2).

c1 = c2 * m2 * (t – t2) / (m1 * (t1 – t)).

Let’s substitute:

c1 = 4200 J / (kg * deg) * 0.4 kg * (25 – 22) / (0.2 kg * (100 – 25) deg) = 336 J / (kg * deg).

Answer: the heat capacity of the metal from which the cylinder is made is 336 J / (kg * deg).