What is the specific heat of a substance from which a metal part weighing 900 g is made, if, being heated to 155 degrees and lowered into a vessel with 3 kg of water at a temperature of 10 degrees, it caused the temperature in the vessel to rise to 15 degrees?
md = 900 g = 0.9 kg.
t1 = 155 ° C.
t2 = 10 ° C.
mw = 3 kg.
t = 15 ° C.
Cw = 4200 J / kg * ° C.
The amount of heat Q during cooling
When a heated part is lowered into water, it cools down from a temperature t1 to a temperature t. The part gives up a certain amount of heat Q to the water, which heats up.
The amount of heat Qd given by the part is found by the formula: Qd = Sd * md * (t1 – t), where
SD – specific heat capacity of the part;
md is the mass of the part;
t1, t – start and end temperature of the part.
The amount of heat when heating
When interacting with a heated part, the water is heated from temperature t2 to temperature t.
The amount of heat Qw, which is necessary for heating water, is determined by the formula: Qw = Sv * mw * (t – t2), where
Sv – specific heat capacity of water;
mw is the mass of heated water;
t1, t – initial and final water temperature.
Heat balance equation
The amount of heat that the part gives off during cooling, Qd, is used to heat the water Qw.
Let’s write down the heat balance equation:
Qd = Qw.
Sd * md * (t1 – t) = Sv * mw * (t – t2).
Let us express the specific heat capacity of the part: SD = Sv * mw * (t – t2) / md * (t1 – t).
Substitute the numerical data from given into the expression.
Сд = 4200 J / kg * ° C * 3 kg * (15 ° C – 10 ° C) / 0.9 kg * (155 ° C – 15 ° C) = 500 J / kg * ° C.
Answer: the specific heat capacity of the substance from which the part is made is
SD = 500 J / kg * ° C.