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 0 degree C (we will allow the ice to sit out for a bit of time and "dry it" so we will make this assumption). For this Pre Lab 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 25 degree C 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 Delta H fusion of ice = 6.0 kj/mol, determine the theoretical final temperature of the water left in the calorimeter when the system reaches equilibrium. degree C Tries 0/99

Explanation / Answer

Heat gained by ice= heat lost by water

m1= mass of ice L=enthalpy of fusion= 6KJ/mol=334J/gK

m2= mass of water

Cp= specific heat capacity of water Let T be the final temperature of the water and ice at equilibrium

m1L+m1Cp*T= m2Cp*(25-T)

on substituting the values in the above equation and solving for T we will get T=7.28degC

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