What amount of heat is absorbed when heating a steel part m = 10.0 kg from a temperature

What amount of heat is absorbed when heating a steel part m = 10.0 kg from a temperature t1 = 100 C to a temperature t2 = 150 C?

m = 10 kg.

t1 = 100 ° C.

t2 = 150 ° C.

C = 500 J / kg * ° C.

Q -?

The amount of thermal energy that is absorbed by a steel part during heating is expressed by the formula: Q = C * m * (t2 – t1), where C is the specific heat capacity of the substance from which the part is made, m is the mass of the part, t2, t1 are the final and initial temperatures details.

Since the part is made of steel, we take the specific heat capacity of steel from the table of specific heat capacity of substances: C = 500 J / kg * ° C.

Q = 500 J / kg * ° C * 10 kg * (150 ° C – 100 ° C) = 250,000 J.

Answer: when heated, a steel part absorbs Q = 250,000 J of thermal energy.