# 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?

## 1 Answer

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).

However, you did make an error in your work, in that you assumed that the ice was solid, but already at 0 degrees Celcius. Because your ice is at -20 degrees Celcius, some of the heat lost by the water will first have to raise the temperate of the ice to 0 degrees Celcius, and then the rest of the heat lost by the water will melt the ice.

• 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
• Looks good. Have fun calculating! – Sean Jan 17 '15 at 3:17