How many degrees could 1 kg of water be heated with the energy needed to melt 1 kg of ice at 0 degrees C?

univerkov | December 2, 2020 | education

Given:
s (c) = 4200 J / kg * ° C
t1 = 0 ° С
m (in) = 1 kg
m (l) = 1 kg
λ (l) = 330,000 J / kg
Find: Δt =?
Decision:
Q (in) = Q (l)
Q (l) = λ * m (l)
Q (l) = 330,000 * 1 = 330,000 J
Q (c) = 330,000 J
Q (in) = with (in) * m (in) * Δt
Δt = Q / c * m
Δt = 330,000 / 1 * 4200 = 78.5 ° С
Answer: Δt = 78.5 ° С

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