1) You cool a 130.0g slug of red-hot iron (temperature 745 ?C) by dropping it in
ID: 2285080 • Letter: 1
Question
1) You cool a 130.0g slug of red-hot iron (temperature 745 ?C) by dropping it into an insulated cup of negligible mass containing 85.0g of water at 20.0 ?C. Assume no heat exchange with the surroundings.
A) What is the final temperature of the water?
B) What is the final mass of the iron and the remaining water?
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2)`An ice cube tray of negligible mass contains 0.285kg of water at 17.4?C. How much heat must be removed to cool the water to 0.00 ?C and freeze it?
A) Express your answer in joules.
B) Express your answer in calories.
C) Express your answer in Btu.
Explanation / Answer
Specific heat capacity of iron is .45J/g*C
Specific heat capacity of water is 4.184J/g*C
Let Tf = Final Temperature, then solve your problem
(mass iron x Spec. Heat Cap. x Change in Temp) + (mass H2O x Spec. Heat Cap. x Change in Temp) = 0
0.45J/gC (130g) (Tf - 745*C) + 4.184J/gc (85g) (Tf - 20*C) = 0
0.00346 (Tf - 745*C) + 0.05 (Tf - 20*C) = 0
0.00346Tf - 2.57 + 0.005Tf - 0.1 = 0
0.00846Tf-2.67=0
Tf = 315.6 C Final Temperature
2)
heat required to cool water to 0 C
Q1 = mw *cw*dT = 0.285*4180*(0 -17.4) = - 18157.92 Joule
>>>>>>> -ve sign means heat taken out
heat required to freeze it to ice at 0 C
Q2 = - mw *L = 0.285*334*10^3 = - 80160 Joule
Total heat to be removed is = -Q1-Q2
Q = 98317.92 Joule = 98.32 Kilo joule
in to joule
=98.32 Kilo joule
=983200J
234889.3879310008 Calories
=222.6 btu
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