We will use 39.9 g of ice and add it to 195.51 g of water at 25degreeC. Let\'s a

Question

We will use 39.9 g of ice and add it to 195.51 g of water at 25degreeC. Let's assume that the ice starts at 0degreeC (we will allow the ice to sit out for a bit of time and "dry it" so we will make this assumption). For this PreLab assignment, let's also assume we have a perfect calorimeter (you will judge this assumption in Part 1 of the lab). So, we have "heat lost" = "heat gained" as Cooling the 195.51 g of water originally in the calorimeter from 25degreeC to the final temperature = melting the 39.9 g of ice + raising the temperature of 39.9 g of the melted ice from 0degreeC to the final temperature. Given that DeltaHfusion of ice = 6.0 kJ/mol, determine the theoretical final temperature of the water left in the calorimeter when the system reaches equilibrium. degreeC Tries 0/99

Explanation / Answer

heat gained by ice on melting = 6000*(39.9/18) J

=13300J

heat gained by ice to reach temp t after melting= 39.9*4.18*t

heat lost by water to reach temp t =

195.51*4.18*(25-t)=13300 +39.9*4.18*t

7230.795=984.038*t

t=7.34 C

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