To determine the specific heat capacity of the liquid in the calorimeter, we can use the principles of heat transfer and the relationship between electrical energy and heat energy. Let's break down the problem step by step.
Understanding the Problem
We have a copper calorimeter filled with a liquid, and we know the following:
- Mass of the liquid (m) = 200 g = 0.2 kg
- Initial temperature (T1) = 25 °C
- Final temperature (T2) = 35 °C
- Time (t) = 5 minutes = 300 seconds
- Current (I) = 20.5 A
- Voltage (V) = 5 V
Calculating Heat Supplied to the Liquid
The electrical energy supplied to the heater can be calculated using the formula:
Electrical Energy (Q) = Voltage (V) × Current (I) × Time (t)
Substituting the values:
Q = 5 V × 20.5 A × 300 s = 30750 J
Heat Gained by the Liquid
The heat gained by the liquid can also be expressed in terms of its mass and specific heat capacity:
Q = m × c × ΔT
Where:
- m = mass of the liquid = 0.2 kg
- c = specific heat capacity of the liquid (unknown)
- ΔT = change in temperature = T2 - T1 = 35 °C - 25 °C = 10 °C
Now we can set the two expressions for Q equal to each other:
30750 J = 0.2 kg × c × 10 °C
Solving for Specific Heat Capacity
Rearranging the equation to solve for c gives:
c = 30750 J / (0.2 kg × 10 °C) = 30750 J / 2 kg·°C = 15375 J/(kg·°C)
Considering the Second Scenario
In the second part of the problem, we need to analyze the situation where the heater maintains the temperature of the calorimeter at 32 °C with a current of 7 A at a potential difference of 6 V. The heat supplied in this case can be calculated similarly:
Q' = 6 V × 7 A × t'
Since the temperature is maintained at 32 °C, the heat supplied must equal the heat lost to the surroundings. The heat loss can be expressed as:
Q' = m × c × (32 °C - 25 °C) = 0.2 kg × c × 7 °C
Setting these equal gives us another equation to solve for c, but since we already calculated c from the first scenario, we can confirm its consistency.
Final Calculation
Using the first calculation, we found that the specific heat capacity of the liquid is:
c = 15375 J/(kg·°C)
This value indicates the amount of heat required to raise the temperature of 1 kg of the liquid by 1 °C. This approach illustrates the relationship between electrical energy and thermal energy, showcasing how energy conservation principles apply in calorimetry.
