To solve this problem, we need to analyze the relationship between power input, temperature rise, and the flow rate of the liquid in the continuous flow 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 formula for heat transfer in a calorimeter can be expressed as:
Q = mcΔT
Where:
- Q = heat energy supplied (in joules)
- m = mass flow rate of the liquid (in kg/s)
- c = specific heat capacity (in J/kg·K)
- ΔT = temperature change (in K)
Calculating Heat Transfer
In the first experiment, the power input (P) is 60W, which means:
P = Q/t
Since we are looking at a continuous flow, we can assume that the heat supplied over a certain time period (t) is equal to the power multiplied by that time. For a steady state, the heat supplied equals the heat lost to the surroundings plus the heat used to raise the temperature of the liquid.
Let’s denote the power lost to the surroundings as P_loss. Therefore, we can express the total power input as:
P = P_loss + Q
Analyzing the First Experiment
From the first experiment, we have:
60W = P_loss + mcΔT
Given that the temperature rise (ΔT) is 10K, we can express the heat gained by the liquid as:
Q = mc(10K)
Examining the Second Experiment
In the second experiment, the power is doubled to 120W, and the same temperature rise of 10K is achieved by increasing the flow rate of the liquid to three times faster. This means:
P = P_loss + mcΔT
Now, since the flow rate is three times faster, the mass flow rate (m) will also be three times greater. Therefore, we can express the heat gained in the second experiment as:
Q = (3m)c(10K)
Setting Up the Equations
For the first experiment:
60W = P_loss + mc(10K)
For the second experiment:
120W = P_loss + (3m)c(10K)
Solving the Equations
From the first equation, we can express P_loss:
P_loss = 60W - mc(10K)
Substituting this into the second equation gives:
120W = (60W - mc(10K)) + (3m)c(10K)
Rearranging this, we find:
120W = 60W + (3m)c(10K) - mc(10K)
This simplifies to:
120W - 60W = (3m)c(10K) - mc(10K)
60W = (2m)c(10K)
Finding Power Lost
Now we can substitute back to find P_loss. From the first experiment, we know:
P_loss = 60W - mc(10K)
We can express mc(10K) from the second experiment:
mc(10K) = 30W
Thus, substituting this back into the equation for P_loss gives:
P_loss = 60W - 30W = 30W
Final Answer
Therefore, the power lost to the surroundings in each case is 30W. The correct answer is B. 30W.
