A piece of copper ball of mass 20g at 200°c is placed in a copper calorimeter of mass 60g containing 50g of water at 30°c ignoring heat losses.calculate the final steady temperature of mixture.(specific heat capacity of water=4.2jg-1k-1)
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To calculate the final steady temperature of the mixture, we can use the principle of heat transfer.
Step 1: Calculate the heat gained by the water:
Q_water = mass_water × specific heat capacity_water × temperature change_water
Here, mass_water = 50g, specific heat capacity_water = 4.2 J/g°C, and temperature change_water = (final temperature - initial temperature) = (final temperature - 30°C).
Step 2: Calculate the heat lost by the copper ball:
Q_copper ball = mass_copper ball × specific heat capacity_copper ball × temperature change_copper ball
Here, mass_copper ball = 20g, specific heat capacity_copper ball = unknown, and temperature change_copper ball = (final temperature - initial temperature) = (final temperature - 200°C).
Step 3: Equate the heat gained by the water and the heat lost by the copper ball:
Q_water = Q_copper ball
mass_water × specific heat capacity_water × temperature change_water = mass_copper ball × specific heat capacity_copper ball × temperature change_copper ball
(50g) × (4.2 J/g°C) × ((final temperature - 30°C)) = (20g) × (specific heat capacity_copper ball) × ((final temperature - 200°C))
Step 4: Solve for the specific heat capacity of copper ball and final temperature:
specific heat capacity_copper ball = (50g) × (4.2 J/g°C) × ((final temperature - 30°C)) / [(20g) × ((final temperature - 200°C))]
Now, let's substitute the values and calculate the final temperature:
specific heat capacity_copper ball = (50g) × (4.2 J/g°C) × ((final temperature - 30°C)) / [(20g) × ((final temperature - 200°C))]
Simplifying the above equation will give us the specific heat capacity_copper ball.
Using this value, we can substitute it back in the equation and solve for the final temperature.
To calculate the final steady temperature of the mixture, we need to use the principle of conservation of energy, which states that energy is conserved in a closed system.
First, let's determine the amount of heat gained or lost by each component:
1. Heat gained or lost by the copper ball:
Q1 = mcΔT1
where m is the mass of the copper ball, c is the specific heat capacity of copper, and ΔT1 is the change in temperature of the copper ball.
Given:
- mass of the copper ball (m) = 20g
- specific heat capacity of copper (c) = ?
- change in temperature of the copper ball (ΔT1) = Tfinal - 200°C
The specific heat capacity of copper is typically 0.386 J/g°C.
So, c = 0.386 J/g°C.
2. Heat gained or lost by the copper calorimeter:
Q2 = mcΔT2
where m is the mass of the copper calorimeter, c is the specific heat capacity of copper, and ΔT2 is the change in temperature of the copper calorimeter.
Given:
- mass of the copper calorimeter (m) = 60g
- specific heat capacity of copper (c) = 0.386 J/g°C
- change in temperature of the copper calorimeter (ΔT2) = Tfinal - 30°C
3. Heat gained or lost by the water:
Q3 = mcΔT3
where m is the mass of the water, c is the specific heat capacity of water, and ΔT3 is the change in temperature of the water.
Given:
- mass of the water (m) = 50g
- specific heat capacity of water (c) = 4.2 J/g°C
- change in temperature of the water (ΔT3) = Tfinal - 30°C
Since the system is closed, the total heat gained by the copper ball, copper calorimeter, and water should be zero:
Q1 + Q2 + Q3 = 0
Substituting the formulas for Q1, Q2, and Q3:
mcΔT1 + mcΔT2 + mcΔT3 = 0
Now we can solve for the final temperature (Tfinal).
After substituting the values, the equation becomes:
(20g × 0.386 J/g°C × (Tfinal - 200°C)) + (60g × 0.386 J/g°C × (Tfinal - 30°C)) + (50g × 4.2 J/g°C × (Tfinal - 30°C)) = 0
Simplifying and solving this equation will give us the final steady temperature (Tfinal).