Question

A thermos contains 144 cm^3 of coffee at 95.0 °C. To cool the coffee, you drop two 10.9-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 in degrees Celsius? Treat the coffee as if it were water.

Answer 1

Answer:$T=69.56^{\circ}$

Explanation:

Given

volume of $V=144 cm^3$

mass of coffee$=\rho \times volume =1\times 144=144 gm$

Initial temperature of coffee$=95^{\circ}$

mass of ice cubes =10.9 gm

and latent heat of ice=0.336 MJ

heat absorb by ice lowers coffee temperature

$m_{ice}\times L=m_{coffee}\times c\times (95-T)$

$10.9\times 0.336\times 10^6=144\times 4.184\times (95-T)$

$336=13.21\times (95-T)$

$T=69.56^{\circ}$