A calorimeter has a mass of 100g and a specific heat of 0.900 cal/g˚c and contains 400g of water at 40˚c. when 91g of ice at 0.0˚c is added and completely melted, the temperature of the water is 18.2˚c. calculate the heat fusion of the ice
Please I need a help on this
To calculate the heat fusion of the ice, we need to use the principle of energy conservation. The heat gained by the water and calorimeter must be equal to the heat lost by the ice. The equation we will use to solve for the heat fusion of the ice is:
Heat gained by water + calorimeter = Heat lost by ice
First, let's calculate the heat gained by the water and calorimeter. We will use the formula:
Q = mcΔT
where Q is the heat gained, m is the mass of the water and calorimeter, c is the specific heat of the water and calorimeter, and ΔT is the change in temperature.
Q(water + calorimeter) = (mass of water + calorimeter) * specific heat * ΔT
Q(water + calorimeter) = (400g + 100g) * 0.900 cal/g˚C * (18.2˚C - 40˚C)
Q(water + calorimeter) = 500g * 0.900 cal/g˚C * (-21.8˚C)
Q(water + calorimeter) = -10,890 cal
Next, let's calculate the heat lost by the ice. The heat lost by the ice is equal to the heat fusion of the ice, which is the energy required to melt the ice without changing its temperature. The equation we will use is:
Q(ice) = mass of ice * heat fusion
Q(ice) = 91g * heat fusion
Now, equating the two equations:
Q(water + calorimeter) = Q(ice)
-10,890 cal = 91g * heat fusion
To solve for the heat fusion, divide both sides of the equation by 91g:
heat fusion = -10,890 cal / 91g
heat fusion ≈ -119.78 cal/g
The heat fusion of the ice is approximately -119.78 cal/g. Note that the negative sign indicates that heat is being lost or released by the system.