Question

A person makes a quantity of iced tea by mixing 470 g of hot tea (essentially water) with an equal mass of ice at its melting point. Assume the mixture has negligible energy exchanges with its environment. (a) If the tea's initial temperature is Ti = 85°C, when thermal equilibrium is reached what are the mixture's temperature Tf and (b) the remaining mass mf of ice? If Ti = 68°C, when thermal equilibrium is reached what are (c) Tf and (d) mf? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg.

Answer 1

a)If T is the final temperature of the mixture, the hot tea has given up heat

(470g)(4.186 J/g-deg)(85 - T)

If all the ice has melted, it would have absorbed heat

(470g)(333 J/g) + (470g)(4.186 J/g-deg)(T - 0oC)

Setting these equal and canceling the common factor of 470 g

(333 J/g) + (4.186 J/g-deg)(T - 0oC) = (4.19 J/g-deg)(85 - T)

Divide by 4.186 J/g-deg

(333/4.186) + T = 85 - T

T = 2.72 degree C

b)mass remaining = 0

c)
If T is the final temperature of the mixture, the hot tea has given up heat

(470g)(4.186 J/g-deg)(68 - T)

If all the ice has melted, it would have absorbed heat

(470g)(333 J/g) + (470g)(4.186 J/g-deg)(T - 0oC)

Setting these equal and canceling the common factor of 470 g

(333 J/g) + (4.186 J/g-deg)(T - 0oC) = (4.19 J/g-deg)(68 - T)

Divide by 4.186 J/g-deg

(333/4.186) + T = 68 - T

T = -5.8 degree C
which is not possible, so Tf = 0

d)
(470g)(4.186 J/g-deg)(68 - 0) = 133785 J

This is sufficient heat to melt (133785 J)/(333 J/g)
= 402 g of ice, so the mass of ice remaining is (470 - 402) = 68 g