Question

A 4kg block of ice at -6°C is put in the water (initial temperature 20°C) in an insulated container.

How much water was there initially, if the final temperature of this system after all of the ice melted is 14°C?

Answer in kilograms to the nearest 0.1kg.

following specific heats and latent heats might be useful:

C_water = 4186J/kgC

C_ice = 2090J/kgC

C_steam =1996J/kgC

_Lfusion  = 333000J/kg

_{Lvaporization}=2.25x10⁶J/kg

Answer 1

Given:

$m=4\:\rm kg$

t_1=-6\:^{\circ}\rm C

t_2=20\:^{\circ}\rm C

t_3=0\:^{\circ}\rm C

t_4=14\:^{\circ}\rm C

$c_{\rm ice} = 2090\:\rm J/kg \cdot C$

$c_{\rm water} = 4186\:\rm J/kg \cdot C$

$\lambda = 333000\:\rm J/kg$

$M-?$

The amount of heat obtained by ice:

$$Q_{\rm ice}=Q_1+Q_2+Q_3\\ =c_{\rm ice} m(t_3-t_1)+\lambda m+c_{\rm water}m(t_4-t_3)$$

$$Q_{\rm ice}=2090*4*6+333000*4+4186*4*14\\ =1616576\:\rm J$$

The heat balance equation says

$$Q_{\rm ice}=Q_{\rm water}$$

Hence

$$1616576\:{\rm J}=c_{\rm water}M(t_2-t_4)$$

Finally

$$M=\frac{1616576}{4186*6}=64.36\:\rm kg$$