Help confusion on heat transfer.

[macmac410]

Homework Statement

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

Homework 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)

The Attempt at a Solution

this is my solution on the problem but as my professor check it, it was wrong, the correct answer that my professor given is 16.73kg, i doubted it because i did a lot of research and thinking in this. and I am shy to correct my professor if his given answer is wrong. please help me if my answer is the right one or is it my professor answers. i have an exam tomorrow and i can't seem to get the right answers unless if my answers is true please correct me if i have miscalculate something or overlooked somthing at my given solution and answers. Thanks you very much. please help me

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

m=0.01089 grams x 1000 = 10.89kg

 

[SteamKing]

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.)