Three hundred grams of ice at 0 degree C is added to 200 grams of water at 95 de

Question

Three hundred grams of ice at 0 degree C is added to 200 grams of water at 95 degree C. The system is kept thermally insulated from its environment. (a) How many joules of energy will it take to melt the ice? (b) What is the equilibrium temperature if the ice in (a) is placed to the water in part (b)? What phases are present? Ten grams of molten lead at 480.0 degree C are added to a water bath originally at 0.00 degree C. The system is kept thermally insulated from its environment so no energy is lost. (c) What is the minimum amount of water needed to cool the lead to a temperature of 99.0 degree C?

Explanation / Answer

(a) to melt the ice,

Q = m Lf

Q = 300 x 334 J/g = 100200 J

(b) lets first check, how much energy can be removed from water as its temp gets to 0 deg C

Q = m C delatT

= 200 x 4.184 x (95 - 0)

= 79496 J

that is less then the energy needed to melt all of the ice.

hence at equiibrium, there will be both ice and water.

eq. temp = 0 deg C

(c) Energy released by lead = 10 x 0.128 x (480 - 99)

= 487.68 J

from energy conservation,

487.68 = m x 4.184 x(99 - 0)

m = 1.18 grams

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