How much heat is required to heat a 20-gram steel needle by 120 degrees?
| November 19, 2020 | education
Given:
Specific heat capacity of steel C = 500 J / kg ° C
mass m = 20 gr = 0.020 kg
t = 120 °
Find: Q
Decision:
Q = C * m * t = 500 * 0.020 * 120 = 1200 J = 1.2 kJ
Answer: Q = 1.2 kJ
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