# Homework Help: Help! confusion on heat transfer.

1. Aug 15, 2011

### macmac410

1. The problem statement, all variables and given/known data
Calculate the amount (in grams) of ice at 0 degree celsius that must be added to an insulated cup with 250g of water at 40 degree celsius to cool the water to 35 degree celsius. ignore heat transfer to the cup.

given data:
T(initial) water= 40 celsius
T(initial) ice = 0
T(final for both h20 and melted ice) = 35

latent heat of fusion of h20 = 334x10^3 j/kg
Heat capacity of h20 = 4190 j/kgC

2. Relevant equations

Summation of heat = Heat ice + heat h20 (breaking down heat of ice = latent heat of ice + sensible heat of ice)

Summation of heat = sensible heat of melted ice + latent heat of ice + sensible heat of h2o at 40)

Qsensible=m(Heat capacity)(deltaT)
Qlatent=m(latent heat of Fusion)

3. The attempt at a solution

the sum of heat is equal to zero since we will ignore the transfer of heat to its environment we will just consider the transfer from h2o to ice.

Summation of heat = sensible heat of melted ice + latent heat of ice + sensible heat of h2o at 40)

0=m(Heat capacity)(deltaT)melted ice+m(latent heat of Fusion) ice + m(Heat capacity)(deltaT)h2o

transpose Q of h2o

-[m(Heat capacity)(deltaT)h2o]= m(Heat capacity)(deltaT)melted ice + m(latent heat of Fusion) ice

factor out mass that is the same on ice

-[m (Heat capacity)(deltaT)h2o]= m [(Heat capacity)(deltaT)melted ice + (latent heat of Fusion) ice]

equate to get the mass

m =-[m(Heat capacity)(deltaT)h2o] / [(Heat capacity)(deltaT)melted ice + (latent heat of Fusion) ice]

substitute the given

m= - [(0.25)(4190)(35-40)]/ [(4190)(35-0)+(334x10^3)]

m=0.01089 grams x 1000 = 10.89kg

Last edited: Aug 15, 2011
2. Aug 15, 2011

### SteamKing

Staff Emeritus
According to your calculations (and your professor's), if I want to cool a glass of water a few degrees, I should use a block of ice about 40 times the mass of the glass of water. I think you and your professor need to put on your thinking caps and recheck your units.

(Ahh, the simplicity of the metric system.)