How Much Ice Melts When Cooling an Enzyme Sample in an Ice Bath?

[winston0064]

Problem: A glass vial containing a .016kg sample of an enzyme is cooled in an ice bath. The bath contains water and .12kg of ice. The sample has specific heat capacity 2250 J/kg k; the glass vial has mass .006kg and specific heat capacity 2800 J/kg K. How much ice melts in cooling the enzyme sample from 19.5 C to the temp of the ice bath?

My work thus far: (initial approach may be what I am finding trouble with...)

Qh2o = .12kg (4190J/kg K)(19.6 - 0 C)
Qenz = .016(2250)(19.6 - 0 C )
Qvial = .006(2800)(19.6 - 0 C)
I believe this problem may involve latent heat but I am stuck as to where to go from here to find the mass of the ice that melted. Any help would would be much appreciated.

 

[HallsofIvy]

You have already calculated the total heat the various things must lose in order to go down to 0 degrees. Yes, you need the "latent heat of melting" (which is exactly the same as the "latent heat of freezing"!). That should be given as the heat required to melt 1 kg of ice. Divide the total heat by that to find out how many kg of ice it can melt.

 

[M98Ranger]

To find the mass of ice that melts, we need to consider the energy required to melt the ice. This can be calculated using the latent heat of fusion, which is the amount of energy required to change a substance from solid to liquid without changing its temperature. For water, the latent heat of fusion is 334 kJ/kg.

First, we need to determine the final temperature of the enzyme sample after it has been cooled in the ice bath. We can use the principle of conservation of energy to set up the equation:

Qh2o + Qenz + Qvial = Qmelt

Where Qh2o is the energy transferred from the water in the ice bath, Qenz is the energy transferred from the enzyme sample, Qvial is the energy transferred from the glass vial, and Qmelt is the energy required to melt the ice.

We can plug in the values we already have for Qh2o, Qenz, and Qvial:

.12kg (4190J/kg K)(Tf - 0 C) + .016(2250)(Tf - 19.6 C) + .006(2800)(Tf - 19.6 C) = Qmelt

Where Tf is the final temperature of the enzyme sample. We can solve for Tf by rearranging the equation and solving for Tf:

Tf = (Qmelt + 0.12kg(4190J/kg K)(0 C) + .016(2250)(19.6 C) + .006(2800)(19.6 C)) / (.12kg(4190J/kg K) + .016(2250) + .006(2800))

Now, we can use this value of Tf to calculate the energy required to melt the ice:

Qmelt = (.12kg)(334kJ/kg) + (.016kg)(2250J/kg K)(Tf - 0 C)

We can plug in the value of Tf we just calculated and solve for Qmelt:

Qmelt = (.12kg)(334kJ/kg) + (.016kg)(2250J/kg K)((Tf - 0 C) - 19.6 C)

Now, we can calculate the mass of ice that melts by dividing Qmelt by the latent heat of fusion:

mice = Qmelt / 334kJ/kg

I hope this helps you to solve the problem.