To determine the specific heat capacity of a substance, a body weighing 400 g, heated to a temperature of 100

To determine the specific heat capacity of a substance, a body weighing 400 g, heated to a temperature of 100, was lowered into a calorimeter containing 200 g of water. The initial temperature of the calorimeter and water was 23. After the thermal equilibrium was established, the temperature of the body and water became equal to 30. Determine the specific heat capacity of the substance of the investigated body. Ignore the heat capacity of the calorimeter.

mt = 400 g = 0.4 kg.

tт = 100 ° C.

mw = 200 g = 0.2 kg.

tv = 23 ° C.

t = 30 ° C.

Cw = 4200 J / kg * ° C.

St -?

The amount of thermal energy Qt, which is released during the cooling of the body, can be expressed by the formula: Qt = St * mt * (tt – t), where St is the specific heat capacity of the body, mt is the body weight, tt, t are the final and initial body temperature.

The amount of thermal energy Qw, which is necessary for heating water, can be expressed by the formula: Qw = Sv * mw * (t – tw), where Sv is the specific heat capacity of water, mw is the mass of water, tw, t are the initial and final water temperatures.

The heat balance equation will have the form: Qt = Qw.

St * mt * (tt – t) = Sv * mw * (t – tv).

St = Sv * mw * (t – tv) / mt * (tt – t).

St = 4200 J / kg * ° C * 0.2 kg * (30 ° C – 23 ° C) / 0.4 kg * (100 ° C – 30 ° C) = 210 J / kg * ° C.

Answer: the specific heat of the body is St = 210 J / kg * ° C.