How much heat is required to raise the temperature of 67.95 grams of PCl3 from 18.6oC to 79.1°C ? Cp ( PCl3) -Specific Heat Capacity= 0.874 J/g°C
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follow the units... you have grams, Temp, and J/gC, and your looking for joules looking for Joules
79.1-18.6 = 60.5
(60.5 C)*(67.95 g)*(.874J/gC) units cancel out except J = 3593 Joules
Sig Figs = 3, so 3590 Joules
79.1-18.6 = 60.5
(60.5 C)*(67.95 g)*(.874J/gC) units cancel out except J = 3593 Joules
Sig Figs = 3, so 3590 Joules
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It's not possible to answer the question from the information you're given.
Your instructor probably just expects you to multiply mass x ΔT x Cp
but the temperature range given includes a phase change.
(MP = -93.6 and BP = 76.1)
heat of temp change (18.6 to 76.1) = m x ΔT x SHC = 67.95g x 57.5° x 0.874J/g° =
heat of phase change (liquid → gas) = m x heat of vaporization. = 67.95g x heat of vaporization
heat of temp change (76.1 to 79.1) = 67.95g x 3° x SHCgas
add these three answers together to get the answer.
Your instructor probably just expects you to multiply mass x ΔT x Cp
but the temperature range given includes a phase change.
(MP = -93.6 and BP = 76.1)
heat of temp change (18.6 to 76.1) = m x ΔT x SHC = 67.95g x 57.5° x 0.874J/g° =
heat of phase change (liquid → gas) = m x heat of vaporization. = 67.95g x heat of vaporization
heat of temp change (76.1 to 79.1) = 67.95g x 3° x SHCgas
add these three answers together to get the answer.
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use this equation:
heat enegy =m x c x change in temperature
=67.95 x0.874 x 60.5
=3592.99215J
heat enegy =m x c x change in temperature
=67.95 x0.874 x 60.5
=3592.99215J