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A small block of copper with a specific heat of 200 cal/0C is initially at a temperature of 3000C. It is then dropped into a large tank of water maintained at 270C. What is the change in entropy of the combined system after equilibrium is established?
- a)Decreases by 100 cal/K
- b)Increases by 100 cal/K
- c)Decreases by 130 cal/K
- d)Increases by 130 cal/K
Correct answer is option 'C'. Can you explain this answer?
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Change in Entropy of the System
Entropy change is given by the equation:
ΔS = Q/T
where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature.
In this scenario, the system consists of the small block of copper and the tank of water. Initially, the block of copper is at a temperature of 30°C, and the water in the tank is at a temperature of 27°C.
Heat Transfer
When the block of copper is dropped into the water, heat transfer occurs between the two substances until they reach thermal equilibrium. The heat transferred can be calculated using the equation:
Q = mcΔT
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Since we are only interested in the change in entropy, we can assume that the specific heat capacity of the water remains constant at 1 cal/g°C.
Using the given values, the heat transferred from the block of copper to the water can be calculated as:
Q = (mcΔT) = (m * 200 cal/°C * (27°C - 30°C)
Calculating the Change in Entropy
Since the heat transferred occurs between two substances, the change in entropy of the system can be calculated as the sum of the change in entropy of the block of copper and the change in entropy of the water.
ΔS_system = ΔS_copper + ΔS_water
The change in entropy of the copper block can be calculated as:
ΔS_copper = Q_copper / T_copper
where Q_copper is the heat transferred from the copper block and T_copper is the temperature of the copper block.
Similarly, the change in entropy of the water can be calculated as:
ΔS_water = Q_water / T_water
where Q_water is the heat transferred to the water and T_water is the temperature of the water.
Substituting the values into the equations, we can calculate the change in entropy of the system:
ΔS_system = (Q_copper / T_copper) + (Q_water / T_water)
Since the temperature of the copper block is higher than the temperature of the water, the heat transferred from the copper block to the water is negative. This means that the change in entropy of the copper block is negative.
However, the heat transferred to the water is positive, resulting in a positive change in entropy for the water.
Therefore, the change in entropy of the system is negative, indicating a decrease in entropy.
Answer: The change in entropy of the combined system decreases by 130 cal/K.
Entropy change is given by the equation:
ΔS = Q/T
where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature.
In this scenario, the system consists of the small block of copper and the tank of water. Initially, the block of copper is at a temperature of 30°C, and the water in the tank is at a temperature of 27°C.
Heat Transfer
When the block of copper is dropped into the water, heat transfer occurs between the two substances until they reach thermal equilibrium. The heat transferred can be calculated using the equation:
Q = mcΔT
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Since we are only interested in the change in entropy, we can assume that the specific heat capacity of the water remains constant at 1 cal/g°C.
Using the given values, the heat transferred from the block of copper to the water can be calculated as:
Q = (mcΔT) = (m * 200 cal/°C * (27°C - 30°C)
Calculating the Change in Entropy
Since the heat transferred occurs between two substances, the change in entropy of the system can be calculated as the sum of the change in entropy of the block of copper and the change in entropy of the water.
ΔS_system = ΔS_copper + ΔS_water
The change in entropy of the copper block can be calculated as:
ΔS_copper = Q_copper / T_copper
where Q_copper is the heat transferred from the copper block and T_copper is the temperature of the copper block.
Similarly, the change in entropy of the water can be calculated as:
ΔS_water = Q_water / T_water
where Q_water is the heat transferred to the water and T_water is the temperature of the water.
Substituting the values into the equations, we can calculate the change in entropy of the system:
ΔS_system = (Q_copper / T_copper) + (Q_water / T_water)
Since the temperature of the copper block is higher than the temperature of the water, the heat transferred from the copper block to the water is negative. This means that the change in entropy of the copper block is negative.
However, the heat transferred to the water is positive, resulting in a positive change in entropy for the water.
Therefore, the change in entropy of the system is negative, indicating a decrease in entropy.
Answer: The change in entropy of the combined system decreases by 130 cal/K.
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A small block of copper with a specific heat of 200 cal/0C is initially at a temperature of 3000C. It is then dropped into a large tank of water maintained at 270C. What is the change in entropy of the combined system after equilibrium is established?a)Decreases by 100 cal/Kb)Increases by 100 cal/Kc)Decreases by 130 cal/Kd)Increases by 130 cal/KCorrect answer is option 'C'. Can you explain this answer?
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