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

Answer:

Mass of water 2.9g

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

$Q_{ice}=Q_{water}$

$1000=340m_{water}$

$m_{water}=2.9g$