How much heat is required to heat a 200 g lead part at 70 ° C.
| February 16, 2021 | education
m = 200 g = 0.2 kg.
Δt = 70 ° C.
C = 140 J / kg * ° C
Q -?
The amount of heat Q, which is necessary for heating a substance, is determined by the formula: Q = C * m * Δt, where C is the specific heat capacity of the substance, m is the mass of the substance, Δt is the change in the temperature of the substance.
We take the specific heat capacity of lead from the table of heat capacity of substances: C = 140 J / kg * ° C.
Q = 140 J / kg * ° C * 0.2 kg * 70 ° C = 1960 J.
Answer: to heat a lead part, Q = 1960 J of thermal energy is needed.
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