Stir 0.25 kg of ice, initially at - 5^{o}C, into 0.5 kg of tea, initially at 45^{o}C.What is the final temperature?c_{w}= 4186 J/(kg C^{o})c

_{i}= 2090 J/(kg C^{o})L

_{f}= 3.33 x 10^{5}J/kgIt seems reasonable to expect all the

iceto melt and to have a final temperature between 0^{o}C and 45^{o}C. We will simply call this final temperature T_{f}. Then thechangein temperature for theteaisT _{1}= T_{f}- T_{1i}

_{}Anything =changein anything = (final value) - (initial value)T

_{1}= T_{f}- 45^{o}CThe heat transferred from the

tea(orbythetea) isQ _{1}= c_{w}m_{1}T_{1}Q

_{1}= [4186 J/(kg C^{o})] [0.5 kg] [T_{f}- 45^{o}C]Notice that [T

_{f}- 95^{o}C] will benegativewhich means Q_{1}will benegative.That means that Q_{1}represents heat flowingfromorout ofthetea.Q _{1}= [4186 J/(kg C^{o})] [0.5 kg] [T_{f}- 45^{o}C]Q

_{1}= 2093 T_{f}J - 94,185 JI have dropped the C

^{o}in the first term, requiring that we measure T_{f}in units of C^{o}.What about the

changein temperature for theice? What about theheatabsorbed by the ice?First the

icechanges temperature from - 5^{o}C to 0^{o}C or a change of T_{2a }= 5^{o}C -- asice! That involves heat ofQ _{2a}= c_{2}m_{2}T_{2a}Q

_{2a}= [2090 J/(kg C^{o})] [0.25 kg] [0^{o}C - ( - 5^{o}C)]Q

_{2a}= [2090 J/(kg C^{o})] [0.25 kg] [5^{o}C]Q

_{2a}= 2612.5 JThis is the heat required to raise the ice to a temperature of 0

^{o}C.Now

allof the ice melts. That requires heat ofQ _{2b}= m_{2}L_{f}Q

_{2b}= [0.25 kg] [3.33 x 10^{5}J/kg]Q

_{2b}= 8.33 x 10^{4}J = 83,300 JNow, this 0.25 kg of water at 0

^{o}C is raised in temperature to the final temperature T_{f}; that requires heat ofQ _{2c}= c_{w}m_{2}T_{2}Q

_{2c}= [4186 J/(kg C^{o})] [0.25 kg] [T_{f}- 0^{o}C]Q

_{2c}= [1046.5 J/C^{o}] [T_{f}]Thus, the

totalheat absorbed by the ice isQ _{2}= Q_{2a}+ Q_{2b}+ Q_{2c}Q

_{2}= 2,612.5 J + 83,300 J + 1046.5 T_{f}JAgain, I have dropped the C

^{o}in the last term, requiring that we measure T_{f}in units of C^{o}.Now we can apply the Work-Energy Theorem or say that energy is conserved. This means

Q _{1}+ Q_{2}= 0[2093 T

_{f}J - 94,185 J] + [2,612.5 J + 83,300 J + 1046.5 T_{f}J] = 0[2093 T

_{f}- 94,185 ] + [2,612.5 + 83,300 + 1046.5 T_{f}] = 03139.5 T

_{f}- 8,272.5 = 03139.5 T

_{f}= 8,272.5T

_{f}= 8,272.5 /3139.5T

_{f}= 2.6Of course, this means

T_{f}= 2.6^{o}C

(c) 2000, Doug Davis; all rights reserved.