A 40-g block of ice is cooled to ?73 Question Part Points Submissions Used Quest

Question

A 40-g block of ice is cooled to

?73

Question Part Points Submissions Used Question Part Points Submissions Used A 40-g block of ice is cooled to -73 degree C and is then added to 600 g of water in an 80-g copper calorimeter at a temperature of 28 degree C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. (If not all the ice melts, determine how much ice is left.) Remember that the ice must first warm to 0 degree C, melt, and then continue warming as water. (The specific heat of ice is 0.500 cal/g degree C = 2,090 3/kg degree C.) Tf = degree C mice final = g

Explanation / Answer

For the ice to reach 0 degrees, you will need 73 * 2090 * 0.04 J = 6102.8 J

The latent heat needed to melt the ice is 334 J / g.
So to melt the ice, you will need 334 * 40 = 13360 J
Assume final temperature = x degC
Specific heat of water is 4.18 J / g degC.
So heat applied to ice will be x * 4.18 * 40 = 1672 x

Total heat energy applied = 6102.8 + 13360 + 1672x

Heat from water = (28 - x) * 4.18 * 600 = 70224 - 2508x

Specific heat of copper is 0.385 J / g degC.

Heat from copper = (28 - x) * 0.385 * 80 = 862.4 - 30.8x

Heat gained = heat loss
6102.8 + 13360 + 1672x = 70224 - 2508x + 862.4 - 30.8x
4210.8 x = 51623.6
x = 12.26oC

Thus final temperature is 12.26oC

Eice = 6102.8 + 13360 + 1672*12.26 = 39961.2 J
Eice = mice * cice * dT
39961.2 = mice * 2090 * 73
mice = 0.262 kg
mice = 262 g

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