At a local convenience store, you purchase a cup of coffee, but, at 98.4°C, it i
ID: 575153 • Letter: A
Question
At a local convenience store, you purchase a cup of coffee, but, at 98.4°C, it is too hot to drink. You add 36.2 g of ice that is-2.2°C to the 248 mL of coffee. What is the final temperature of the coffee? (Assume the heat capacity and density of the coffee are the same as water and the coffee cup is well insulated.) This is all the information that was given in the problem. ): View comments (1)> Expert Answer Anonymous answered this 25 answers Was this answer helpful?o1 Heat edl hise heht lost by Caffee 244632-79-GY = 28.2 T 24323·Explanation / Answer
It is given that the heat capacity and the density of water are the same as that of ice. The density of water is 1 g/mL. Assume the heat capacity of ice to be S J/g.°C.
Let the constant temperature reached be t°C. The coffee is at a higher temperature; hence, it will lose heat. The heat lost by coffee = (mass of coffee)*S*(temperature change of coffee) = (248 mL)*(1 g/mL)*S*(98.4 – t)°C = 248*S*(98.4 – t) g.°C.
The ice will gain heat, since it is at a lower temperature; however, note that at 0°C, the ice will melt to form liquid water. Hence, we need to consider the fusion of ice as well.
The heat gained by ice can be broken up into three parts:
(i) heat gained by ice to reach 0°C = (36.2 g)*S*[(0°C) – (-2.2°C)] = 79.64*S g.°C.
(ii) heat required to melt ice = (mass of ice)*(latent heat of fusion of ice) = (36.2 g)*(333.55 J/g) = 12074.51 J.
(iii) heat required by ice to reach t°C = (36.2 g)*S*(t – 0)°C = 36.2*S*t g.°C.
By the problem,
79.64*S g.°C + 12074.51 J + 36.2*S*t g.°C = 248*S*(98.4 – t) g.°C
We know that S = 4.18 J/g.°C; plug in values and obtain
79.64*(4.18 J/g.°C)g.°C + 12074.51 J + 36.2*(4.18 J/g.°C)*t g.°C = 248*(4.18 J/g.°C)*(98.4 – t) g.°C
====> 332.8952 J + 12074.51 J + 151.316*t J = 1036.64*(98.4 – t) J
====> 12407.4052 J + 151.316*t J = 102005.376 J – 1036.64*t J
====> 151.316*t + 1036.64*t = 102005.376 – 12407.4052
====> 1187.956*t = 89597.9708
====> t = 89597.9708/1187.956 = 75.4219
The temperature attained by the coffee is 75.4219°C 75.42°C (ans).
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