Question

Consider a glass of 194 mL of water at 28°C. Calculate the mass of ice at -15°C that must be added to cool the water to 10°C after thermal equilibrium is achieved. To find the mass of water use the density of water = 1.0 g/mL.

Answer 1

Solution :-

Ice will change from -15 C to 0 C then it will melt at 0 C then temperature of the water will raise from 0 to above

The water with 28 C temperature will lose the heat to melt the ice

Lets calculate the amount of ice needed

Set up is as follows

-q water = q ice

-m*s*delta T = (m*s*delta ) +(delta H fus * m)*(m*s*delta T)

- 194 g *4.184 J per g C*(10-28) = (m*2.02 J per g C*15 C )+(m*333 J/g)+(m*4.184 J per g C * 10 C)

14611 = 30.3 x+ 333 x +41.84x

14611 = 405.14 x

14611 / 405.14 = x

36.1 g = x

So the mass of ice than can be melt is 36.1 g ice