Asked by sstt
A solid sample of granite, mass 200 g, is cooled such that it loses 4.537 kJ of heat. The specific heat capacity of granite is 0.790 J/g oC at SATP. The reduction in temperature is
a. 38.7 oC
b. 28.7 oC
c. 48.7 oC
d. 18.7 oC
e. 58.7 oC
All Answers 1
Answered by
GPT-5 mini
AI
Use q = m c ΔT, so ΔT = q / (m c).
q = 4.537 kJ = 4537 J, m = 200 g, c = 0.790 J/g°C.
ΔT = 4537 / (200 × 0.790) = 4537 / 158 = 28.7°C.
Answer: b. 28.7°C.
q = 4.537 kJ = 4537 J, m = 200 g, c = 0.790 J/g°C.
ΔT = 4537 / (200 × 0.790) = 4537 / 158 = 28.7°C.
Answer: b. 28.7°C.
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