Specific Heat. Copper has a specific heat of 0.092 cal/g/deg C. A 75 gram copper mass has been heated to a temperature of 100 degrees C. The copper is placed in thermal contact with 50 grams of water at 25 degrees C. What will be the final equilibrium temperature of the copper and water? (Specific heat of water = 1 cal/g/deg C.)
Sum of heats gained=0 (some lose heat, so are negative)
Heatgained water+heatgainedCu=0
50*1*(tf-25)+ 75*.092*(tf-100)=0
solve for tf.
To find the final equilibrium temperature, we can use the principle of heat transfer and the concept of specific heat.
1. Calculate the heat absorbed by the copper:
Q1 = m1 * c1 * ΔT1
where m1 is the mass of copper (75g), c1 is the specific heat of copper (0.092 cal/g/deg C), and ΔT1 is the change in temperature for the copper (final temperature - initial temperature).
ΔT1 = 100°C - initial temperature
2. Calculate the heat released by the water:
Q2 = m2 * c2 * ΔT2
where m2 is the mass of water (50g), c2 is the specific heat of water (1 cal/g/deg C), and ΔT2 is the change in temperature for the water (final temperature - initial temperature).
ΔT2 = initial temperature - 25°C
3. Since the system is in thermal equilibrium, the heat gained by the water is equal to the heat lost by the copper:
Q1 = Q2
4. Substituting the values into the equation, we get:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
5. Rearranging the equation to solve for the final temperature of the system:
ΔT1 / ΔT2 = (m2 * c2) / (m1 * c1)
final_temperature - initial_temperature / initial_temperature - 25 = (50 * 1) / (75 * 0.092)
6. Simplifying the equation:
final_temperature - 100 / initial_temperature - 25 = 50 / (75 * 0.092)
final_temperature - 100 / initial_temperature - 25 = 0.907
7. Cross-multiplying and rearranging the equation:
final_temperature - 100 = 0.907 * (initial_temperature - 25)
final_temperature - 100 = 0.907 * initial_temperature - 22.675
8. Bringing all the terms involving final_temperature to one side and all the constant terms to the other side:
final_temperature - 0.907 * initial_temperature = -22.675 + 100
final_temperature - 0.907 * initial_temperature = 77.325
9. Solving for the final_temperature:
final_temperature = 0.907 * initial_temperature + 77.325
Finally, you can plug in the value of the initial temperature to find the final equilibrium temperature of the copper and water.
To find the final equilibrium temperature of the copper and water, we need to use the principle of conservation of energy. The heat gained by the water must be equal to the heat lost by the copper.
First, let's calculate the heat gained by the water. We can use the formula:
Qwater = mass x specific heat x change in temperature
Qwater = 50 g x 1 cal/g/deg C x (final temperature - initial temperature)
= 50 cal/deg C x (final temperature - 25 deg C)
Now, let's calculate the heat lost by the copper. We can use the same formula:
Qcopper = mass x specific heat x change in temperature
Qcopper = 75 g x 0.092 cal/g/deg C x (100 deg C - final temperature)
Since the heat gained by the water must be equal to the heat lost by the copper, we can set up an equation:
Qwater = Qcopper
50 cal/deg C x (final temperature - 25 deg C) = 75 g x 0.092 cal/g/deg C x (100 deg C - final temperature)
Let's solve this equation to find the final equilibrium temperature.
50 final temperature - 50 x 25 = 75 x 0.092 (100 - final temperature)
50 final temperature - 1250 = 6.9 (100 - final temperature)
50 final temperature - 6.9 final temperature = 1250 + 6.9 x 100
43.1 final temperature = 1250 + 690
43.1 final temperature = 1940
final temperature = 1940 / 43.1
final temperature ≈ 45.04 deg C
So, the final equilibrium temperature of the copper and water will be approximately 45.04 degrees Celsius.