Can someone please help me with steps?
2018-01-16 7:11 pm
A 42.02 g sample of a substance is initially at 24.9 °C. After absorbing 2105 J of heat, the temperature of the substance is 122.1 °C. What is the specific heat (c) of the substance?
回答 (2)
2018-01-16 7:35 pm
For the sample of a substance :
Heat absorbed, Q = 2105 J
Mass, m = 42.02 g
Specific heat, c = ? J/g°C
Raise in temperature, ΔT = (122.1 - 24.9)°C = 97.2°C
Heat absorbed, Q = m c ΔT
Specific heat of the substance, c = Q / (m ΔT) = 2105 / (42.02 × 97.2) J/g°C = 0.515 J/g°C
Heat absorbed, Q = 2105 J
Mass, m = 42.02 g
Specific heat, c = ? J/g°C
Raise in temperature, ΔT = (122.1 - 24.9)°C = 97.2°C
Heat absorbed, Q = m c ΔT
Specific heat of the substance, c = Q / (m ΔT) = 2105 / (42.02 × 97.2) J/g°C = 0.515 J/g°C
2018-01-16 7:35 pm
heat energy = mass X specific heat X change in temp
2105 = 42.02 X SH X ( 122.1 - 24.9)
SH = 2105 / ( 42.02 X 97.2 )
SH = 0.515J/g°C
2105 = 42.02 X SH X ( 122.1 - 24.9)
SH = 2105 / ( 42.02 X 97.2 )
SH = 0.515J/g°C
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