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

A cooking vessel on a slow burner contains 10.0 kg of liquid water and an unknown mass of ice in equilibrium at 0ºC at time t=0.00 minutes. The temperature of the mixture is measured at various times and the result is plotted below. During the first 50.0 minutes, the mixture remains at 0ºC. From 50.0 minutes until 60.0 minutes, the temperature increases linearly to 2.00ºC. The specific heat capacity of liquid water is 4190 j/kg•K. The latent heat of fusion of ice is 333x10^3J/kg. Assuming that the burner supplies heat at a constant rate, and neglecting the specific heat capacity of the vessel, please find the initial mass of the ice.

Graph that's posted is exactly as the problem states, Flat line at 0.0 ºC measured every minute for 50 minutes, then a rise of .2ºC each minute until 2.0ºC at 60 mins is reached.

2. Relevant equations

Q=mLf

Q=mc(delta)T

dQ/dt=mcdT/dt --?(this is where i'm not sure if I should use it or not)

3. The attempt at a solution

I know the mass in the 2nd part of the equation will be 10.0kg+m(ice), using the dQ equation I end up having my dts cancel, but I know it has to be a time dependent problem or they wouldn't have given us the time. When I set up

dT/dt = (dQ/dt)/mc and put dT=.2 and dt=60s, they cancel and I'm left with Q=(10+m(ice))*(838). There's got to be different calculus that I'm not seeing here, unless I really just plug that Q in to the inititial Q=mLf equation and get a mass, but that just seems too simple.

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# Thermodynamics, heat of fusion

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