J.R. S. answered • 10/10/19
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q = mCΔT
q = heat =?
m = mass = 472 ml x 1g/ml = 472 g
C = specific heat = 4.184 J/g/deg (assuming the same C as water)
ΔT = change in temperature = 35 degrees
q = (472g)(4.184 J/g/deg)(35 deg) = 69,120 j
69,120 J x 1 cal/4.184 J = 16,520 cal = 16.5 kcal
Kade H.
What about this: How many Cals are required to heat a 28.4 g ice cube from -23.0 °C to -1.0 °C.
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10/10/19
J.R. S.
tutor
q = mCdeltaT
Look up the specific heat for ice (H2O solid) and plug it in.
q = 28.4 g x Cice x 22 deg
Easiest if you use Cice in cal. If you use joules, convert the answer as done in the original question, i.e 1 cal = 4.184 J
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10/10/19
Kade H.
Thank you, you are the freakin man!10/10/19