A100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is added to the water and ice in the container. what is the final temperature of the system? (specific heat of ice = 2,000 j/kg°c , specific heat of water = 4,186 j/kg°c, heat of fusion of water = 333.7 kj/kg, specific heat of glass = 837.2 j/km°c, specific heat of lead = 127.7 j/km°c)

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

Let the final temperature of the system be . Thus

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable .

Confirm the uniformity of units, equate the two expressions and solve for :

which goes against the initial assumption. Implying that the final temperature does not go above the melting point of water- i.e., . However, there's no way for the temperature of the system to go below ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

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