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

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SK93 SK93 · Answer 1 · 2019-12-21T11:20:58+01:00

Answer:

Mass of water 2.9g

Explanation:

Ice

[tex]m_{ice}=100g[/tex]

[tex]c_{ice}=2J/g.K[/tex]

[tex]T_{ice,initial}=-10\°C[/tex]

[tex]T_{ice,final}=T_{equilibrium}=-5\°C[/tex]

Water

[tex]c_{water}=4J/g.K[/tex]

[tex]T_{water,initial}=10\°C[/tex]

[tex]T_{water,final}=0\°C[/tex]

[tex]T_{equilibrium}=-5\°C[/tex]

[tex]l_{water}=300J/g[/tex]

[tex]m_{water}=?g[/tex]

Step 1: Determine heat gained by ice

[tex]Q_{ice}=m_{ice}c_{ice}(T_{ice,final}-T_{ice,initial})[/tex]

[tex]Q_{ice}=100*2*(-5--10)[/tex]

[tex]Q_{ice}=1000J[/tex]

Step 2; Determine heat lost by water

[tex]Q_{water}=m_{water}c_{water}(T_{water,initial}-T_{water,final})+m_{water}l_{water}[/tex]

[tex]Q_{water}=m_{water}*4*(10-0)+m_{water}*300[/tex]

[tex]Q_{water}=40m_{water}+300m_{water}[/tex]

[tex]Q_{water}=340m_{water}[/tex]

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

[tex]Q_{ice}=Q_{water}[/tex]

[tex]1000=340m_{water}[/tex]

[tex]m_{water}=2.9g[/tex]