Question

A thermos contains 150 cm3 of coffee at 85 ????C. To cool the coffee, you drop two 11-g ice cubes into the thermos. The ice cubes are initially at 0 ????C and melt completely. What is the final temperature of the coffee? Treat the coffee as if it were water.

Answer 1

Answer:

73.73°C is the final temperature of the coffee.

Explanation:

Heat lost by the coffee = Q

Mass of the coffee = m

Volume of coffee = V = $150 cm^3=150 mL (1cm^3=1 mL)$

Density of water = Density of coffee solution = d = 1 g/mL (given)

$Mass =density\times volume$

$m=1 g/mL\times 150 mL=150 g$

Heat capacity of the coffee is equal to that of water= c = 4.18 J/g°C

Initial temperature of the coffee = $T_1=85^oC$

Final temperature of the coffee = T

$Q=mc\times (T-T_1)$

Heat required to melt 11 grams of melt ice = Q'

Latent heat of ice = $\Delta H_{lat}=334 J/g$

$Q'=334J/g\times 11g=3674 J$

Heat absorbed by the ice after melting = q

Mass of ice melted into water = m' = 11 g

Heat capacity of water = c = 4.18 J/g°C

Initial temperature of water =$T_2$ = 0°C

Final temperature of water = T

$q=m'\times c\times (T_2-T)$

According law of conservation of energy , energy lost by coffee will equal to heat required to melt ice and further to raise the temperature of water.

$-Q=Q'+q$

$-(mc\times (T-T_1))=m'\times c\times (T-T_2)+3674 J$

$150 g\times 4.18 J/g^oC\times (85^oC-T)=11 g\times 4.18 J/g^oC\times (T-0^oC)+3674 J$

On solving we get:

T = 73.73°C

73.73°C is the final temperature of the coffee.