(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

Q=(L)(m)

3. 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???

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# An Ice Cube is Added to a Thermos of Coffee

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