# An Ice Cube is Added to a Thermos of Coffee

## Homework Statement

An insulated Thermos contains 110 cm3 of hot coffee at 87.0°C. You put in a 15.0 g ice cube at its melting point to cool the coffee. By how many degrees (in Celsius) has your coffee cooled once the ice has melted? Treat the coffee as though it were pure water and neglect energy exchanges with the environment. The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg. The density of water is 1.00 g/cm3.

Q=(m)(c)(Tf-Ti)
Q=(L)(m)

## The Attempt at a Solution

I tried a couple of things. I am confident I could do the question if I didn't have to deal with the latent heat of fusion. I made two equations:
q= (mass of cube)(specific heat)(273.15-Tf)
q= (mass of water)(specific heat)(Tf-360.15)
and for Q of melting, I got 4.995 kJ. I'm just not sure how to incorporate that in. I tried adding it to the specific heat in the first equation and I tried multiplying it in, but neither approach worked. Suggestions Please???

## Answers and Replies

Sorry....it seems like some people have viewed, but haven't posted yet. Is there anything I did wrong or should add? Basically, I'm just trying to figure out how the latent heat of fusion is incorporated into the two equations I set up above.

HallsofIvy
Science Advisor
Homework Helper
You have calculated the energy used in melting the ice cube (energy of fusion). Now you need to calculate two more things. What is the energy required to raise the 15g= 15cc ice water produced by the ice cube to "T" degrees (for unknown T)? What is the energy released as the 110 cc of coffee cools from 87 to T degrees? Both of those, of course, depend on T.

Energy of fusion+ energy to raise the icewater to T= energy released by cooling the coffee to T.

Solve that equation for T.

Ok. I'm still not getting the right answer for some reason! I have for my expression (Lm)+mc(T-273.15)=-mc(360.15-T)
What am I missing now?