A 0.0500-kg ice cube at −30.0°C is placed in 0.450 kg of 35.0°C water in a very well-insulated container. What is the final temperature in degrees Celsius?
Given
mass of the ice cubes is m1 = 0.0500
kg
initial temperature is T1 = -30.0 0C
mass of water i s m2 = 0.450 kg
initial Temperature of the water is T2 = 35.0 0C
final temperature of the system is T = ?
here the heat energy lost by the ice cubes = heat energy
gain by the water
and the ice will change the state at 0 0C so
the equation can be written as
m1*C1*(0-(-30))+m1*L+m1*C2(T-0) =
m2*C2*(T-T2)
here C1 is specific heat of ice C1 = 2010 J/kg.k
latent heat of fusion of ice is L = 333*10^3 J/kg
specific heat of water is C2 = 4184 J/kg.k
so
0.0500*2010(30)+0.0500*333*10^3+0.0500*4184(T-0) =
0.450*4184(35-T)
solving for T , T = 22.1 0C
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