A sample of an unknown substance has a mass of 0 point 1 5 8 kilograms If 2 comma 510 point 0 cap j of heat is required to heat the substance from 32 point 0 times degree C to 61 point 0 times degree C , what is the specific heat of the substance? Use q is equal to m cap c sub p triangle cap t
Answer
A
15 point 9 times cap j over g times degree cap c
B
0 point 1 7 1 times cap j over g times degree cap c
C
0 point 5 4 8 times cap j over g times degree cap c
D
86 point 6 times cap j over g times degree cap c
Explanation
Use q = mcₚΔT and solve for specific heat: cₚ = q/(mΔT). Convert 0.158 kg to 158 g, and find the temperature change: 61.0 − 32.0 = 29.0°C. Then cₚ = 2,510 J ÷ (158 g × 29.0°C) ≈ 0.548 J/(g·°C), so the specific heat is choice C.
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