A 35.8g sample of a substance is initially at 27.8 degree C. After absorbing 1543 J of heat, the temperature of the substance is 106.0 C.?

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[匿名]

2016-02-22 7:27 pm

What is the specific heat (c) of the substance?

回答 (1)

andrew

2016-02-22 7:36 pm

Heat absorbed = m c ΔT

1543 = 35.8 × c × (106.0 - 27.8)

Specific heat, c = 1543 / [35.8 × (106.0 - 27.8)]

c = 0.551 J/(g °C)

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