Initially 48g of ice at 0 degrees celsius is in an aluminium calorimeter can of mass 2.0 g also at 0 degrees celsius.Then 75g OF water at 80 degrees celsius is poured into the can.What is the final temperature.
...aluminium calorimeter can of mass 2.0 g(!)
Is it possible?
To find the final temperature after mixing the ice and water, we can use the principle of conservation of energy. The heat lost by the water when it cools down will be equal to the heat gained by the ice when it melts and heats up.
The equation we can use is:
(heat lost by water) = (heat gained by ice)
The heat lost by the water can be calculated using the equation:
Q = m * c * ΔT
Where:
Q is the heat lost by the water
m is the mass of the water
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature
Since the water cools down from 80°C to the final temperature, we can write the equation as:
Q = 75g * 4.18 J/g°C * (80°C - final temperature)
The heat gained by the ice can be calculated using the equation:
Q = m * ΔHf + m * c * ΔT
Where:
Q is the heat gained by the ice
m is the mass of the ice (48g)
ΔHf is the heat of fusion of water (334 J/g)
c is the specific heat capacity of ice (2.09 J/g°C)
ΔT is the change in temperature
Since the ice melts at 0°C and heats up to the final temperature, we can write the equation as:
Q = 48g * 334 J/g + 48g * 2.09 J/g°C * final temperature
Now, equating the heat lost by the water to the heat gained by the ice:
75g * 4.18 J/g°C * (80°C - final temperature) = 48g * 334 J/g + 48g * 2.09 J/g°C * final temperature
Simplifying the equation will give us the final temperature.