Question

A 40-g block of ice is cooled to

−69°C

and is then added to 570 g of water in an 80-g copper calorimeter at a temperature of 23°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
mice final	=	____________g

Answer 1

Assuming all the ice is melt and final temperature of system is T, then if final temperature comes negative, than some there must be some amount of ice remaining, And if final answer is positive, than that temperature is final temperature of system and mass of ice will be 0 gm

Suppose the final temperature is T.

Now Using energy conservation:

Heat released by water and calorimeter = Heat gained by ice

Q1 + Q2 = Q3 + Q4 + Q5

m1*C1*dT1 + m2*C2*dT2 = m3*C3*dT3 + m3*Lv + m3*C1*dT4

Q1 = Heat released by water from 23 C to T C = m1*C1*dT1

Q2 = Heat released by calorimeter from 23 C to T C = m2*C2*dT2

dT1 = dT2 = 23 - T

m1 = 570 gm = 0.57 kg = mass of water

m2 = 80 gm = 0.08 kg = mass of calorimeter

C1 = Specific heat capacity of water = 4186 J/kg-C

C2 = Specific heat capacity of copper = 387 J/kg-C

Q3 = Heat gained by ice from -69 C to 0 C = m3*C3*dT3

Q4 = Heat gained during phase change = m3*Lf

Q5 = Heat gained by ice from 0 C to T C = m3*C1*dT4

dT3 = 0 - (-69) = 69 C

dT4 = T - 0 = T C

C3 = specific heat capacity of ice = 2090 J/kg-C

Lf = 3.33*10^5 J/kg = latent heat of fusion

m3 = mass of total ice = 40 gm = 0.04 kg

Now using given values:

0.57*4186*(23 - T) + 0.08*387*(23 - T) = 0.04*2090*69 + 0.04*3.33*10^5 + 0.04*4186*T

Now Solving above equation

T = (0.57*4186*23 + 0.08*387*23 - 0.04*2090*69 - 0.04*3.33*10^5)/(0.57*4186 + 0.08*387 + 0.04*4186)

T = 14.12 C = 14 C

Since T > 0, So final amount of ice = 0 gm

Please Upvote.