A 400-g block of copper at a temperature of 85°C is dropped into 300 g of water at 33°C. The water is contained in a 250-g glass container. What is the final temperature of the mixture?
The sum of heats gained is zero.
masswater*c*(Tf-33)+masscopper*c.cu*(Tf-85)=0
solve for Tf
To find the final temperature of the mixture, we can use the principle of conservation of energy. The total amount of heat gained by the water and the glass container should be equal to the total amount of heat lost by the copper block.
We can calculate the amount of heat gained by the water using the formula:
Qwater = mwater * cwater * ΔTwater
Where:
Qwater is the heat gained by the water
mwater is the mass of the water (300 g)
cwater is the specific heat capacity of water (4.18 J/g°C)
ΔTwater is the change in temperature of the water
We can calculate the amount of heat gained by the glass using the formula:
Qglass = mglass * cglass * ΔTglass
Where:
Qglass is the heat gained by the glass
mglass is the mass of the glass container (250 g)
cglass is the specific heat capacity of the glass (0.84 J/g°C)
ΔTglass is the change in temperature of the glass container
We can calculate the amount of heat lost by the copper using the formula:
Qcopper = mcopper * ccopper * ΔTcopper
Where:
Qcopper is the heat lost by the copper
mcopper is the mass of the copper block (400 g)
ccopper is the specific heat capacity of copper (0.39 J/g°C)
ΔTcopper is the change in temperature of the copper block
Since the final temperature of the mixture will be the same for both the water and the glass container, we can set Qwater + Qglass = Qcopper.
By rearranging the equation and solving for ΔTwater:
ΔTwater = (Qcopper - Qglass) / (mwater * cwater + mglass * cglass)
Substituting the given values:
ΔTwater = (Qcopper - Qglass) / (300 g * 4.18 J/g°C + 250 g * 0.84 J/g°C)
Now, let's calculate each term separately:
Qcopper = mcopper * ccopper * ΔTcopper
Qcopper = 400 g * 0.39 J/g°C * (Tfinal - 85°C)
Qglass = mglass * cglass * ΔTglass
Qglass = 250 g * 0.84 J/g°C * (Tfinal - 33°C)
Substituting these values back into the equation for ΔTwater:
ΔTwater = (400 g * 0.39 J/g°C * (Tfinal - 85°C) - (250 g * 0.84 J/g°C * (Tfinal - 33°C))) / (300 g * 4.18 J/g°C + 250 g * 0.84 J/g°C)
Now, we can solve this equation to find the value of ΔTwater, which will give us the final temperature of the mixture.