55) A 200-g block of copper at a temperature of 90 C is dropped into 400g of water at 27 C. The water is contained in a 300-g glass container. What is the final temperature of the mixture?
I will be happy to check your thinking. This is mostly an algebra problem.
Yes I know but I am confused on how to separate T(final) out of the equation. I can arrange it fine but don't know how to solve for T(Final)
To find the final temperature of the mixture, we can use the principle of heat transfer, which states that the heat gained by one object is equal to the heat lost by another object.
First, let's find the heat lost by the copper block. We can use the specific heat capacity formula:
Q1 = m1 * c1 * ΔT1
Where:
Q1 is the heat lost by the copper block
m1 is the mass of the copper block (200 grams)
c1 is the specific heat capacity of copper (0.39 J/g°C)
ΔT1 is the change in temperature of the copper block (final temperature - initial temperature)
Next, let's find the heat gained by the water and the glass container. We can use the same formula:
Q2 = (m2 + m3) * c2 * ΔT2
Where:
Q2 is the heat gained by the water and glass container
m2 is the mass of the water (400 grams)
m3 is the mass of the glass container (300 grams)
c2 is the specific heat capacity of water (4.18 J/g°C)
ΔT2 is the change in temperature of the water and glass container (final temperature - initial temperature)
Since heat is being transferred from the copper block to the water and glass container, we can set Q1 equal to Q2:
m1 * c1 * ΔT1 = (m2 + m3) * c2 * ΔT2
Rearranging the equation to solve for the final temperature:
ΔT2 = (m1 * c1 * ΔT1) / ((m2 + m3) * c2)
Final temperature = initial temperature + ΔT2
Now let's plug in the given values:
m1 = 200g
c1 = 0.39 J/g°C
ΔT1 = initial temperature - final temperature = 90°C - final temperature
m2 = 400g
m3 = 300g
c2 = 4.18 J/g°C
Final temperature = 27°C + [(m1 * c1 * (90°C - final temperature)) / ((m2 + m3) * c2)]
Solving this equation will give us the final temperature of the mixture.