After 6.00 kg of water at 79.9 oC is mixed in a perfect thermos with 3.00 kg of

Question

After 6.00 kg of water at 79.9 oC is mixed in a perfect thermos with 3.00 kg of ice at 0.0 oC, the mixture is allowed to reach equilibrium. When heat is added to or removed from a solid or liquid of mass m and specific heat capacity c, the change in entropy can be shown to be S = mc ln(Tf/Ti), where Ti and Tf are the initial and final Kelvin temperatures. Using this expression and the change in entropy for melting, find the change in entropy that occurs.

*Note: First, use the fact that the heat gained by the ice is equal to the heat lost by the water to solve for the equilibrium temperature. The specific heat capacity of water is 4186 J/(kg K) and the latent heat of fusion of water is 3.35 x 105 J/kg. Be careful with the units of temperature.

Explanation / Answer

Mass of ice = 3*1000 = 3000 g
Mass of Water = 6 * 1000 = 6000 g

Let the final Temperature be Tf

Heat Gained by ice =
Q = mice * Lice + mice * cwater * T
Q = c0 + 3000 * 4.186 * Tf

Heat lost by Water =
Q = mwater * c water * T
Q = 6000 * 4.186 * (79.9 - Tf)

Heat Gained by ice = Heat lost by Water
3000 * 335.0 + 3000 * 4.186 * Tf = 6000 * 4.186 * (79.9 - Tf)
Tf = 26.59 o C

Final Temperature in Kelvin = 299.75 K
Change in entropy, S = mc ln(Tf/Ti)

S = 6000 * 4.186 * ln(299.75/353)
S = 6000 * 4.186 * -0.1635
S = - 4106.46 J K1
As the temperature of the water decreased, the entropy decreased.

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