A calorimeter of water equivalent 50g contains 250g of water and 50g of ice at 0 degree celsius, 30 g of water at 80 degree celsius is added to it. The final condition in the system will be
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The final condition in the system will be 280g of water at a temperature of 40 degrees Celsius.
To determine the final condition of the system, we need to consider the heat transfer between the substances.
1) Heat transfer from the hot water:
The heat transfer equation is given by Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
The heat transferred from the hot water (30 g at 80 degrees Celsius) to the cold substances (250 g of water and 50 g of ice at 0 degrees Celsius) can be calculated as follows:
Q_hotwater = m_hotwater * c_water * ΔT_hotwater
Q_hotwater = 30 g * 4.18 J/g°C * (0 - 80) °C
Q_hotwater = -9960 J
(Note: The negative sign indicates heat loss from the hot water.)
2) Heat transfer to the ice:
The heat transfer equation for phase change is given by Q = m * L, where Q is the heat transferred, m is the mass, and L is the latent heat of fusion for ice (334 J/g).
The heat transferred to the ice can be calculated as follows:
Q_ice = m_ice * L
Q_ice = 50 g * 334 J/g
Q_ice = 16700 J
3) Heat transfer to raise the temperature of ice to 0 degrees Celsius:
The heat transfer equation is the same as mentioned earlier (Q = mcΔT).
The heat transferred to raise the temperature of the ice to 0 degrees Celsius can be calculated as follows:
Q_ice_temp = m_ice * c_ice * ΔT_ice_temp
Q_ice_temp = 50 g * 2.09 J/g°C * (0 - (-10)) °C
Q_ice_temp = 1045 J
(Note: The negative sign indicates heat loss.)
Now, let's calculate the total amount of heat transferred:
Total Q = Q_hotwater + Q_ice + Q_ice_temp
Total Q = -9960 J + 16700 J + 1045 J
Total Q = 7685 J
Since the system is adiabatic (no heat exchange with the surroundings), the final condition will be that the combined mass of the water and ice will have a final temperature between 0 and 10 degrees Celsius. The specific final temperature cannot be determined without knowing the specific heat capacity of the calorimeter and its initial temperature.
To determine the final condition of the system, we need to consider the principle of conservation of energy.
First, let's calculate the heat required to melt the ice in the calorimeter. The heat required to melt the ice can be calculated using the formula:
Q = m * L
Where:
Q is the heat required,
m is the mass of the ice,
L is the latent heat of fusion of ice.
Given that the mass of ice is 50g and the latent heat of fusion of ice (L) is 334 J/g, we can calculate the heat required to melt the ice.
Q = 50g * 334 J/g = 16,700 J
Next, let's calculate the heat gained by the water at 80 degrees Celsius when it is mixed with the rest of the system. The heat gained can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained,
m is the mass of the water at 80 degrees Celsius,
c is the specific heat capacity of water,
ΔT is the change in temperature.
Given that the mass of the water at 80 degrees Celsius is 30g, the specific heat capacity of water (c) is 4.18 J/g°C, and the change in temperature (ΔT) is the final temperature of the system minus 80 degrees Celsius. We can assume the final temperature is the same for both the ice and water.
Let's denote the final temperature of the system as T. The heat gained by the water at 80 degrees Celsius can be calculated as:
Q = 30g * 4.18 J/g°C * (T - 80)
Now, let's consider the heat exchanged between the ice, water, and the calorimeter. The heat gained by the ice will be equal to the heat lost by the water at 80 degrees Celsius and the calorimeter during the process of reaching a thermal equilibrium. We can use the equation:
Q_ice = Q_water + Q_calorimeter
Given that the water equivalent of the calorimeter is 50g, we can calculate the heat gained by the water equivalent using:
Q_calorimeter = m_calorimeter * c * ΔT_calorimeter
Where:
m_calorimeter is the mass of the water equivalent (which is 50g in this case),
c is the specific heat capacity of water,
ΔT_calorimeter is the change in temperature of the water equivalent (final temperature minus 0 degrees Celsius).
Substituting the given values, we have:
Q_calorimeter = 50g * 4.18 J/g°C * T
Considering that the heat gained by the ice is equal to the heat lost by the water at 80 degrees Celsius and the calorimeter, we have:
Q_ice = 16,700 J = Q_water + Q_calorimeter
Substituting the calculated values, we get:
16,700 J = 30g * 4.18 J/g°C * (T - 80) + 50g * 4.18 J/g°C * T
Now, we can solve this equation to find the final temperature (T).