# Calculate mass of ice needed to cool water $\Delta$T degrees

I have a question about a thermodynamics formula I'd like to apply in my solution to a problem.

The problem is "Determine how much ice one needs to toss into boiling water of mass $m_{w}$ such that the ice completely melts and the water cools down to $0^{\circ}C$. The ice has initial temperature $-20^{\circ}C$. Assume the specific heats of ice and water are $c_{i}$ and $c_{w}$ respectively, and that the latent heat of melting ice is $L$.

My solution is to use the formula for the heat required for phase change: $\Delta Q =L\Delta m$ and equate the exchanged heat with the formula for heat exchange: $\Delta Q=cm\Delta T$.

So $Lm_{i}=c_{w}m_{w}\Delta T$ and $m_{i}=\dfrac{c_{w}m_{w}}{L}\Delta T=100\dfrac{c_{w}m_{w}}{L}$.

My question is if I correctly used $\Delta Q=cm\Delta T$. Is $m_{w}$ the correct variable to use, since the mass of water exchanging heat is $m_{w}$? Additionally, is the equality between the two equations I have used valid for this problem?

Assuming your problem is a closed system, your heat lost by the boiling water will be equal to the heat gained by the ice, given by the equation $Q_{lost}=Q_{gained}$. The mass of the ice term, $m_i$ will appear only on the $Q_{gained}$ side of the equation and the mass of the water term, $m_w$ will appear only on the $Q_{lost}$ side of the equation (at least until you begin to rearrange terms).
• Using the equation $Q_{lost}=Q_{gained}$ we have $c_{w}m_{w}(100^{\circ}C)=c_{i}m_{i}(20^{\circ}C)+Lm_{i}$ which, isolating $m_{i}$ gives, $m_{i}=\dfrac{c_{w}m_{w}(100^{\circ})}{20^{\circ}Cc_{i}+L}$. Is that what you mean? – user44816 Jan 17 '15 at 2:46