Answer:
The heat generated by the electric heater can be calculated using the formula:
Q = I * V * t
where Q is the heat generated (in joules), I is the current (in amperes), V is the voltage (in volts), and t is the time (in seconds).
In this case, the voltage applied is 30 V, the current observed is 2.5 A, and the time is 3.0 min = 180 s. Thus, the heat generated is:
Q = 2.5 A * 30 V * 180 s = 13,500 J
The calorimeter has a mass of m = 100 g = 0.1 kg, and the temperature increase is ΔT = 25.59°C - 20.00°C = 5.59°C. The specific heat capacity of water is c = 4.18 J/g°C.
The heat absorbed by the calorimeter can be calculated using the formula:
Q = m * c * ΔT
Substituting the known values gives:
13,500 J = 0.1 kg * 4.18 J/g°C * 5.59°C
Solving for the specific heat capacity of the calorimeter C gives:
C = 13,500 J / (0.1 kg * 5.59°C * 4.18 J/g°C) ≈ 63.5 J/°C
Therefore, the specific heat capacity of the calorimeter is approximately 63.5 J/°C.
Step-by-step explanation: