The temperature of a 34.93 mg sample of a newly created insulating material rose

Question

The temperature of a 34.93 mg sample of a newly created insulating material rose from 280.6 K to 504.9 K when 68.8 J of heat was applied. What is the specific heat (in J/(ºC•g) of this material? Do NOT show units in your answer.
(in J/(ºC•g)

Please show me how you did the steps. I'm having a hard time figuring it out. Thank you!

The temperature of a 34.93 mg sample of a newly created insulating material rose from 280.6 K to 504.9 K when 68.8 J of heat was applied. What is the specific heat (in J/(ºC•g) of this material? Do NOT show units in your answer.
(in J/(ºC•g)

Explanation / Answer

Q = m * c * delta t

68.8 = 0.03493*C*(504.9 - 280.6)

68.8 = 0.03493 *C*(224.3)

68.8 = 7.8348C

C = 68.8/7.8348 = 60.9652

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