0.7 kg of ice taken at 0 degrees was turned into water at 45 degrees. How much heat was required for this?
| December 3, 2020 | education
Given:
m = 0.7kg
t1 = 0 ° C
t2 = 45 ° C
λ = 330,000 J / kg
c = 4200 J / (kg * ° C)
Find: Q -?
Q = λ * m + c * m * (t2-t1) = 330,000 * 0.7 + 4200 * 0.7 * (45-0) = 363300 (J) = 363.3 (kJ)
Answer: Q = 363.3 kJ.
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