posted by anthony
A 40-g block of ice is cooled to
and is then added to 590 g of water in an 80-g copper calorimeter at a temperature of 27°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°C, melt, and then continue warming as water. (The specific heat of ice is 0.500 cal/g · °C = 2,090 J/kg · °C.)
Tf = ? (°C)
mass of ice final (g)
0 degree celcius because all the ice has not melted or 4 degree celcius.
For the ice to reach 0 degrees, you will need Q = c• m• ΔT 0.04• 2090 • 71 = 5935,6 J
The latent heat needed to melt the ice is 334 000 J / kg.
So to melt the ice, you will need 334 000 • 0.040 = 13360 J
Assume final temperature = x degC
Specific heat of water is 4180 J / k g degC.
So heat applied to ice will be 0.04• 4180• x = 1672 x
Total heat energy applied = 5935,6 + 13360 + 167,2x
Heat from water =0.590 • 4180 • (27-x) = 2466.2• (27-x)
Specific heat of copper is 385 J / kg degC.
Heat from copper = 0.08• 385 • (27-x) = 30.8• (27-x)
Heat gained = heat loss
5935,6 + 13360+167.2 x =2466.2(27-x) +30.8• (27-x)
19296 +167.2 x = 2466.2• (27-x) +30.8• (27-x)
So final temperature is 8.53 degC.