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1. Set up the expressions representing the heat lost and the heat gained, and subsequently derive 3). 2. What does the water equivalent of the calorimeter represent? 3. Set up the expressions for the heat lost and the heat gained, and subsequently derive (4). 4. Before the ice is placed in the calorimeter, the ice is dried. Why did it need to be dry? If the ice was not dry, would the experimental value of the latent heat of fusion be higher or lower than the expected value? Explain. 5. If ice cubes are taken from a freezer and used immediately in the experiment, then will the value of the latent heat of fusion be less than or greater than the theoretical value? Explain. solved for the latent heat of fusion L, the result is L=m_c. (T-T) + mc (T-T)-m,c T m, where cw is the specific heat of water (4.19 J/g C) and mecc, is the water equivalent of the calorimeter. 3-1 In order to calculate the heat of fusion of ice from (3), it is necessary to first determine the water equivalent of the calorimeter. The value mce is found by mixing known quantities of warm water and cool water in the calorimeter. (3) Suppose the Calorimeter contains a mass of warm water mww, at temperature Tw. If a mass of ool water mew, at temperature Te is mixed with the warm water in the calorimeter. thermal equilibrium will be established at an intermediate temperature T When the heat lost and the heat gained are equated, and the resulting equation solved for the water equivalent of the calorimeter. the expression becomes _m_c(T T.) m(T -T) T-T where cw is the specific heat of water.