A glass vial containing a 16.0-g sample of an enzyme is cooled in an ice bath. The bath contains water and?
0.120 kg of ice. The sample has specific heat capacity 2250 J/kg*K ; the glass vial has mass 6.0 g and specific heat capacity 2800 J/kg*K. How much ice (in grams) melts in cooling the enzyme sample from room temperature (19.5°C) to the temperature of the ice bath?
Heat (Q) gained by the ice will melt the ice and is equal to Q=mL where L is the latent heat of fusion of the ice which is around 333 J/g.
Heat (Q) lost by the enzyme and vial will lower their temperature according to Q = mC(change in temp).
The heat gain and heat loss will equal zero when added together since no outside heat is added or removed, so mL + mC(change in temp) + mC(change in temp) = 0
There are two mC(change in temp)'s because both the vial and enzyme lose energy.
Making sure everything's in grams, you get:
m(333 J/g) + 6 g x 2.8 J/g*K x (-19.5K) + 16g x 2.25 J/g*K x (-19.5K) = 0
Solving for m gives 2.4 g of ice will melt.
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