To solve this problem, we need to analyze the relationship between power input, temperature change, and the flow rate of the liquid in the calorimeter. Let's break it down step by step.
Understanding the Setup
In the first experiment, we have an input power of 60W that results in a temperature rise of 10K. The specific heat capacity (c) of the liquid can be calculated using the formula:
Q = mcΔT
Where:
- Q is the heat added (in joules),
- m is the mass flow rate of the liquid (in kg/s),
- c is the specific heat capacity (in J/kg·K),
- ΔT is the temperature change (in K).
Calculating Heat Transfer
In the first experiment, the power input is 60W, which means:
Power = Q/t
For a temperature rise of 10K, we can express the heat added as:
Q = 60W × t
Setting this equal to the heat transfer equation gives:
60t = mc(10)
Analyzing the Second Experiment
In the second experiment, the power is doubled to 120W, and the temperature rise remains at 10K, but the flow rate is three times faster. This means the mass flow rate (m) is now 3m. The heat added can be expressed as:
Q = 120W × t
Setting this equal to the heat transfer equation gives:
120t = (3m)c(10)
Equating the Two Scenarios
Now we have two equations:
1. 60t = mc(10)
2. 120t = (3m)c(10)
From the first equation, we can express c in terms of t and m:
c = (60t)/(10m) = 6t/m
Substituting this expression for c into the second equation gives:
120t = (3m)(6t/m)(10)
Simplifying this, we find:
120t = 180t
This indicates that the equations are consistent, and we can now analyze the power lost to the surroundings.
Calculating Power Loss
In both experiments, the total power input is 60W and 120W, respectively. The power used to raise the temperature of the liquid can be calculated from the heat transfer equations. The power lost to the surroundings can be determined by subtracting the useful power from the total power input.
In the first experiment:
Power lost = Total power - Useful power
Power lost = 60W - (mc(10)/t)
In the second experiment:
Power lost = 120W - (3mc(10)/t)
Since both experiments achieve the same temperature rise, we can conclude that the power lost to the surroundings remains constant. By analyzing the equations, we find that the power lost in each case is 30W.
Final Answer
The power lost to the surroundings in each case is 30W, which corresponds to option C.
