How much heat must be spent to heat 250 kg of steel from 20 to 1020 ° C?

univerkov | November 27, 2020 | education

Condition:
m = 250 kg.
t1 = 20 ° C.
t2 = 1020 ° C.
c = 500 J / kg * ° С.
c-specific heat capacity of steel.
To find :
Q-? J.
Decision :
Q = c * m * Δt.
Q = c * m * (t2-t1).
[Q] = J / kg * ° С * kg * ° С.
Q = 500 * 250 * (1020-20) = 125000000 (J).
Answer: 125 MJ.

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