A piece of copper ball of mass 20g at 200°c is placed in a copper calomiter of mass 60g containing 50g of water at 30°c ignoring heat losses.calculate the final steady temperature of the mixture
You not only want us to do the problem for you, you want me to Google the specific heat of copper and water. LOL
20 Cc (200 - T) = 60 Cc(T-30)+50Cw(T-30)
I need a worked out solution for this question
No
Please solve the above question
To find the final steady temperature of the mixture, we can use the principle of conservation of energy. The heat gained by the water will be equal to the heat lost by the copper ball.
The heat gained or lost can be calculated using the formula:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass
c = specific heat capacity
ΔT = change in temperature
For the water:
m1 = 50g (mass of water)
c1 = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = Tf - 30°C (change in temperature of water)
For the copper ball:
m2 = 20g (mass of copper ball)
c2 = 0.39 J/g°C (specific heat capacity of copper)
ΔT2 = Tf - 200°C (change in temperature of copper ball)
Since the heat gained by the water is equal to the heat lost by the copper ball, we can write the equation:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Substituting the given values:
50g * 4.18 J/g°C * (Tf - 30°C) = 20g * 0.39 J/g°C * (Tf - 200°C)
Now, we can solve for Tf (the final steady temperature of the mixture) by rearranging the equation and solving for Tf:
(50 * 4.18 * Tf) - (50 * 4.18 * 30) = (20 * 0.39 * Tf) - (20 * 0.39 * 200)
(209 * Tf) - (6270) = (7.8 * Tf) - (7800)
209Tf - 6270 = 7.8Tf - 7800
201.2Tf = 7800 - 6270
201.2Tf = 1530
Tf = 1530 / 201.2
Tf ≈ 7.61°C
Therefore, the final steady temperature of the mixture is approximately 7.61°C.