A 100 aluminum calorimeter contains 200g of water with an intial temperature of 15 degree celsius. if a 95g piece of copper heated to a temperatude 90 degree celsius is dropped into water, what will be the final temperature of the mixture? neglect heat loss to the surrounding.
The heat lost by the copper equals the heat gained by the water. Write that in equation form and solve for the final temperature.
19.35 degrees celsius
100
29.0419194
33.03797
To find the final temperature of the mixture, we can use the principle of conservation of energy. The energy gained or lost by one substance is equal to the energy gained or lost by the other substance.
The equation we can use for this calculation is:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0
Where:
m1 = mass of the water (200g)
c1 = specific heat capacity of water (4.18 J/g·°C)
ΔT1 = change in temperature of water (final temperature - initial temperature)
m2 = mass of the copper (95g)
c2 = specific heat capacity of copper (0.38 J/g·°C)
ΔT2 = change in temperature of copper (final temperature - initial temperature)
We need to solve for the final temperature, so we rearrange the equation:
ΔT1 + ΔT2 = 0
To substitute the values into the equation, we need to calculate the energy gained or lost by each substance. The energy gained or lost by a substance can be calculated using the formula:
Q = m * c * ΔT
Where Q is the energy gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
Calculating the energy gained or lost by water:
Q1 = m1 * c1 * ΔT1 = 200g * 4.18 J/g·°C * ΔT1
Calculating the energy gained or lost by copper:
Q2 = m2 * c2 * ΔT2 = 95g * 0.38 J/g·°C * ΔT2
Since we know that the energy gained by one substance is equal to the energy lost by the other substance (according to the principle of conservation of energy), we can set Q1 equal to -Q2.
Therefore,
m1 * c1 * ΔT1 = -m2 * c2 * ΔT2
Substituting the values:
200g * 4.18 J/g·°C * ΔT1 = -(95g * 0.38 J/g·°C * ΔT2)
Simplifying the equation:
4.18 * ΔT1 = -0.38 * ΔT2
Now, we can substitute this simplified equation into the previous equation:
ΔT1 + ΔT2 = 0
4.18 * ΔT1 + ΔT1 = 0
5.18 * ΔT1 = 0
ΔT1 = 0
Now, we substitute the value of ΔT1 = 0 into the simplified equation:
4.18 * 0 = -0.38 * ΔT2
0 = -0.38 * ΔT2
Since the value of ΔT2 cannot be zero, we conclude that there is no change in temperature for the copper. This means that the final temperature of the mixture will be equal to the initial temperature of the water, which is 15°C.