Question

If mr.M has a cup of water with room temperature of 20 °C but adds 80 g of ice to the drink, so it cools to 6°C and adds more ice to drink of water so the temperature is 10°C what will the volume of the drink be?

Answer 1

1.

$$cM(T_w-T_1)=cm(T_1-T_i)+mL$$

$$M(4200)(20-6)=80(4200(6-0)+334000)$$

Initial mass of water:

$$M=489\ g$$

2.

$$c(M+m)(T_2-T_1)=cm'(T_w-T_2)$$

$$(M+m)(T_2-T_1)=m'(T_w-T_2)$$

$$(489+80)(10-6)=m'(20-10)$$

$$m'=228\ g$$

Total mass:

$$M+m+m'=489+80+228=797\ g$$

Total volume is

$$v=0.797\approx0.8\ L$$