A COPPER CALORIMETER WEIGHS 200G WHEN EMPTY AND 400G WHEN HALF-FILLED WITH WATER AT 30DEGREES.10G OF STEAM IS PASSED INTO THE CALORIMETER UNTIL A FINAL STEADY TEMPERATURE IS REACHED.NEGLECTING HEAT LOSSES TO THE SURROUNDING,CALCULATE THE FINAL TEMPERATURE OF THE CALORIMETER AND CONTENT(SPECIFIC LATENT HEAT OF VAPORISATION OF STEAM=2,260,000J/KG)
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To solve this problem, we'll use the principle of energy conservation. The heat gained by the water and calorimeter will be equal to the heat lost by the steam.
Given information:
- Mass of the empty copper calorimeter (m_calorimeter) = 200g
- Mass of the half-filled calorimeter and water (m1) = 400g
- Mass of steam added (m2) = 10g
- Initial temperature of the water and calorimeter (T1) = 30°C
- Specific latent heat of vaporization for steam (L) = 2,260,000 J/kg
Step 1: Calculate the mass of water in the half-filled calorimeter:
m_water = m1 - m_calorimeter = 400g - 200g = 200g
Step 2: Calculate the heat gained by the water and calorimeter:
Q_gain = m_water * c_water * ΔT
Where
- c_water is the specific heat capacity of water, which is approximately 4,186 J/kg°C.
- ΔT is the change in temperature, which is the final temperature (T2) minus the initial temperature (T1).
Step 3: Calculate the mass of the steam:
m_steam = m2 = 10g
Step 4: Calculate the heat lost by the steam:
Q_loss = m_steam * L
Step 5: Set the heat gained by the water and calorimeter equal to the heat lost by the steam:
Q_gain = Q_loss
Step 6: Substitute the values and solve for the final temperature (T2):
m_water * c_water * ΔT = m_steam * L
200g * 4,186 J/kg°C * (T2 - 30°C) = 10g * 2,260,000 J/kg
8,372(T2 - 30) = 22,600
8,372T2 - 250,170 = 22,600
8,372T2 = 272,770
T2 = 32.58°C
Therefore, the final temperature of the calorimeter and its contents is approximately 32.58°C.
To calculate the final temperature of the calorimeter and its contents, we need to use the principle of conservation of energy.
First, let's analyze the energy changes that occur during the experiment:
1. Heat gained by the water: The water initially at 30 degrees Celsius absorbs heat until it reaches the final temperature.
2. Heat lost by the steam: The 10g of steam loses heat until it condenses into water at its boiling point.
3. The final temperature of the system is the temperature at which the heat gained and lost are equal.
Now, let's solve the problem step by step:
Step 1: Calculate the heat gained by the water.
The specific heat capacity of water is 4,186 J/(kg·°C), which means that 1g of water requires 4,186 J to increase its temperature by 1°C.
The mass of water in the calorimeter is half-filled, which is equal to (400g - 200g) = 200g = 0.2kg.
The temperature change for the water is the final temperature minus the initial temperature, which is T - 30.
Therefore, the heat gained by the water is:
Heat_gained = mass_water * specific_heat_water * temperature_change
Heat_gained = 0.2kg * 4,186 J/(kg·°C) * (T - 30)
Step 2: Calculate the heat lost by the steam.
The specific latent heat of vaporization of steam is given as 2,260,000 J/kg.
The mass of the steam is given as 10g = 0.01kg.
Therefore, the heat lost by the steam (energy required for condensation) is:
Heat_lost = mass_steam * latent_heat_vaporization
Heat_lost = 0.01kg * 2,260,000 J/kg
Step 3: Set up the equation and solve for the final temperature.
According to the principle of conservation of energy, the heat gained is equal to the heat lost:
Heat_gained = Heat_lost
Substitute the expressions for heat gained and heat lost into the equation:
0.2kg * 4,186 J/(kg·°C) * (T - 30) = 0.01kg * 2,260,000 J/kg
Simplifying the equation:
4,186 J/(kg·°C) * (T - 30) = 2,260,000 J/kg
Now, solve for T:
(T - 30) = (2,260,000 J/kg) / (4,186 J/(kg·°C))
(T - 30) = 539.53 °C/(kg·°C)
T = 539.53 °C + 30 °C
T = 569.53 °C
Therefore, the final temperature of the calorimeter and its contents is approximately 569.53 °C.