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## Homework Statement

Your 200-g cup of tea is boiling-hot. About how much ice should your add to

bring it down to a comfortable sipping temperature of65°C . Assume that the ice is

initially at−15°C . The specific heat capacity of ice is 0.5cal g⋅°C , for water is 1 cal g⋅°C.

The latent heat for melting ice is 80cal g.

## Homework Equations

Equation for latent heat: L=Q/m

## The Attempt at a Solution

First I need to find the heat lost by the water. This is done using:

Q=c

_{w}m

_{w}ΔT → (1 cal g⋅°C)(200 g)(65 C - 100 C)

Q=-7000 calories

Assuming no heat is lost to anything else during the process, Q

_{lost}= Q

_{gained}

So the ice cube gains the heat lost by the water, or 7000 calories.

Here is where I am stuck.

I tried using the latent heat equation directly (L=Q/m → m=Q/L) using the latent heat of melting ice a 80 cal/g, but this gave me the incorrect answer.

How does one figure out the mass? I tried subbing in the specific heat capacity for Q in the latent heat eqn, but then my masses cancelled. So that didn't work.