## A quick description of your question...

At a local convince store, you purchase a cup of coffe, but at 98.4 degree, it is too hot to drink. You add 47.7g ice that is -2.2 dgree to the 248ml of coffe. What is the final temperatuer of the coffe.( assume that the the heat capacity and dinsity of the coffe are the same as the water. and the coffe coups are well insluated.
• Anonymous commented

I dont understand what u mean. The calculations I have rechecked. I came up with same answer again. The answer might vary a little based on data you have used.

• Anonymous commented

if I give you M.P. 1ATM 0 degree? does will change the answer.

• Anonymous commented

No, It will not. the pressure was already assumed to be constant and 1 atm. and MP 0 C

• Anonymous commented

Well I think you did help me a lot I like to say thank you. Even the computer gave me wrong for the answer. it could be set up mistake.

• let the final temperature be T degree Celcius.

In this case, the ice will undergo 3 stages :

1. Ice at -2.2 degree celcius will go to Ice at 0 degree Celcius.

2. Ice will melt at 0 degree Celcius.

3. Water formed will get heated from 0 degree Celcius to T degree Celcius.

Heat lost by cofee = (m *C* ΔT)coffee = 248* 4.2* (98.4 - T) ---------------------------------------------(1)

Heat gained by ice and water

= (m *C* ΔT)ice from -2.2 to o degree + Latent heat of melting +(m *C* ΔT)water from 0 to T degree celcius

=(47.7*2.108*2.2)+ (47.7*334) + (47.7*4.2*T)

= 16153.01+ 200.34T..............................................................................................(2)

Equating Equations 1 & 2:

248* 4.2* (98.4 - T) = 16153.01+ 200.34T

=> 86340.43= 1241.94 T

=> T= 69.52 degree Celcius

• Anonymous commented

not quit right

• Anonymous commented

do you think if you know M.P. 1 ATM 0 dgree , will help you to answer it

• Anonymous commented

Sorry I just find out the answer was 63, after all you were neare to the answer.

• Heat gained by ice = heat lost by Coffee

For Ice :
Heat gained = m*S ice * 2.2 + m* H fus + m * S water * T

For coffee :
Heat lost = M * S coffee * (98.4 - T) = M * S water * (98.4 - T) {mass of coffee= volume of it, as density is 1g/cm2}

Equating and putting,
S ice = 0.5 cal/g C ; S water = 1 cal/g C ; H fus = 80 cal/g

calculate T = 69.45 C
• Anonymous commented

It seems that should be the answer but , it is chaking it wrong mark. Thank you any way

• The heat loss of the coffee is M(coffee)*cpw*(98.4 - Tf). The heat removed by the ice is 48.2*cpi*2.2 + 48.2*Wf + 48.2*cpw*Tf. Tf is the final temperature, cpw is the heat capacity of water/coffee, cpi is the heat capacity of ice and Wf is the heat of fusion of ice. Equate these

248*cpw*(98.4 - Tf) = 48.2*cpi*2.2 + 48.2*Wf + 48.2*cpw*Tf

Look up cpi, cpw and Wf and solve for Tf
• Anonymous commented

Please I need help little more

• The heat loss of the coffee is M(coffee)*cpw*(98.4 - Tf). The heat removed by the ice is 48.2*cpi*2.2 + 48.2*Wf + 48.2*cpw*Tf. Tf is the final temperature, cpw is the heat capacity of water/coffee, cpi is the heat capacity of ice and Wf is the heat of fusion of ice. Equate these

248*cpw*(98.4 - Tf) = 48.2*cpi*2.2 + 48.2*Wf + 48.2*cpw*Tf

248* 4.18 *10^3*(98.4 - Tf) = 48.2*2.108 *10^3*2.2 + 48.2 * 334 * 10^3 * Tf
102005376-1036640 Tf =223532.32 + 16098800 Tf
so 101781843.68 = 17135440 Tf
Tf = 5.939 C
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