120 grams of boiling water (temperature 100° C, heat capacity 4.2 J/gram/K) are

Question

120 grams of boiling water (temperature 100° C, heat capacity 4.2 J/gram/K) are poured into an aluminum pan whose mass is 1100 grams and initial temperature 22° C (the heat capacity of aluminum is 0.9 J/gram/K).

(a) After a short time, what is the temperature of the water?
Tfinal =    C

(b) What simplifying assumptions did you have to make?
-Energy transfer between the system (water plus pan) and the surroundings was negligible during this time.
-The heat capacities for both water and aluminum hardly change with temperature in this temperature range.
-The thermal energy of the aluminum doesn't change.
-The thermal energy of the water doesn't change.


(c) Next you place the pan on a hot electric stove. While the stove is heating the pan, you use a beater to stir the water, doing 26432 J of work, and the temperature of the water and pan increases to 77.8° C. How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan?
Q =      J

Explanation / Answer

mass of the water m1 = 120g initial temp t1 = 100 deg heat capacity c1 =4.2J/g/K final temp =t loss of heat by water Q1 = m1c1(t1-t) mass of pan m2 = 1100g intial temp t2 = 22 deg heat capacity c2 = 0.9J/g/K final temp =t gain of heat Q2 = m2c2(t-t2) a) if no heat transmits to surroundings in the short time then Q1=Q2 120*4.2*(100-t)=1100*0.9*(t-22) 50400-504t=990t-21780 1494t=72180 t=48.31deg b) we should assume -Energy transfer between the system (water plus pan) and the surroundings was negligible during this time c) on heating the temp raises to 77.8 deg gain by water Q3 = m1c1(77.8-48.31) Q3=120*4.2*29.49 =14862.96J gain by pan Q4 =m2c2(77.8-48.31) Q4=1100*0.9*29.49=29195.1J total gain Q5=Q3+Q4 =44058.06J the energy due to stirrer Q6 = 26432J the energy supplied from stove Q = Q5-Q6 Q=44058.06-26432=17626.06J

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