Thermal Physics

1. The problem statement, all variables and given/known data

0.02 kg of ice and 0.10 kg of water at 0 degrees Celsius are in a container. steam at 100 degrees Celsius is passed in until all the ice is just melted. How much water is now in the container?

Specific latent heat of steam = 2.3 * 10^6 J/kg
Specific latent heat of ice = 3.4 * 10^5 J/kg
Specific heat capacity of water = 4.2 * 10^3 J/kg/K

2. Relevant equations

latent heat Q = mL
Specific Heat Capacity Q = mCT

Where Q = Heat required
m = mass of substance
C = Specific heat Capacity
T = change in Temperature

3. The attempt at a solution

Using latent heat equation, the heat required for a phase change from ice to water

Q = 0.02 (3.4 x 10^5)
= 6800 J

Using Specific heat capacity equation, the heat require for water

Q = 0.10(4.2 x 10^3)100
= 42,000 J

Total heat required

6800 + 42000 = 48,800 J

plugging in 48,800 into latent heat equation for steam

48,800 = m (2.3 x 10^6)
m = 0.0212 kg

will this answer be the mass of the water in the container?
or am i missing steps or is completely wrong with my workings???

http://ift.tt/1p5BQPy

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