Given these data, compute the heat of fusion of water The specific heat capacity of water is 4.186 J/(g.K). Mass of calorimeter = Specific heat of calorimeter = 3.00 x 10 g 0.380 J/(g.K Initial temperature of water and Mass of water = 2.00 x 10 g Mass of ice 30.0 gInitial temperature of ice 0°C calorimeter 20.0°C Final temperature of calorimeter- 8.5°C Solution Heat gained by Ice = Heat lost by calorimeter + water m_Ice*Lf + m_Ice*C_Water*(8.5 - 0) = m_cal*C_cal*(20 - 8.5) + m_water*C_water*(20 - 8.5) 30*Lf + 30*4.186*8.5 = 3*10^2*0.38*11.5 + 2*10^2*4.186*11.5 Lf = (3*10^2*0.38*11.5 + 2*10^2*4.186*11.5 - 30*4.186*8.5)/30 = 329 J/g or 3.29*10^5 J/kg <<<<<<------Answer .

1. Given these data, compute the heat of fusion of water The specific heat capacity of water is
4.186 J/(g.K). Mass of calorimeter = Specific heat of calorimeter = 3.00 x 10 g 0.380 J/(g.K
Initial temperature of water and Mass of water = 2.00 x 10 g Mass of ice 30.0 gInitial
temperature of ice 0°C calorimeter 20.0°C Final temperature of calorimeter- 8.5°C
Solution
Heat gained by Ice = Heat lost by calorimeter + water
m_Ice*Lf + m_Ice*C_Water*(8.5 - 0) = m_cal*C_cal*(20 - 8.5) + m_water*C_water*(20 -
8.5)
30*Lf + 30*4.186*8.5 = 3*10^2*0.38*11.5 + 2*10^2*4.186*11.5
Lf = (3*10^2*0.38*11.5 + 2*10^2*4.186*11.5 - 30*4.186*8.5)/30
= 329 J/g or 3.29*10^5 J/kg <<<<<<------Answer