Question

You place 100 grams of ice, with a temperature of −10∘C, in a styrofoam cup. Then you add an unknown mass of water, with a temperature of +10∘C, and allow the system to come to thermal equilibrium. For the calculations below, use the following approximations. The specific heat of solid water is 2 joules / gram, and the specific heat of liquid water is 4 joules / gram. The latent heat of fusion of water is 300 joules / gram. Assume no heat is exchanged with the surroundings. Express your answers in grams, using two significant digits.

If the final temperature of the system is -5∘C , how much water was added? ______________ grams

Answer 1

Mass of water 2.9g

Step-by-step explanation:

Ice

`m_(ice)=100g`

`c_(ice)=2J/g.K`

`T_(ice,initial)=-10\°C`

`T_(ice,final)=T_(equilibrium)=-5\°C`

Water

`c_(water)=4J/g.K`

`T_(water,initial)=10\°C`

`T_(water,final)=0\°C`

`T_(equilibrium)=-5\°C`

`l_(water)=300J/g`

`m_(water)=?g`

Step 1: Determine heat gained by ice

`Q_(ice)=m_(ice)c_(ice)(T_(ice,final)-T_(ice,initial))`

`Q_(ice)=100*2*(-5--10)`

`Q_(ice)=1000J`

Step 2; Determine heat lost by water

`Q_(water)=m_(water)c_(water)(T_(water,initial)-T_(water,final))+m_(water)l_(water)`

`Q_(water)=m_(water)*4*(10-0)+m_(water)*300`

`Q_(water)=40m_(water)+300m_(water)`

`Q_(water)=340m_(water)`

Step 3: Heat gained by ice is equivalent to heat lost by water

`Q_(ice)=Q_(water)`

`1000=340m_(water)`

`m_(water)=2.9g`