The stored electrical energy of a 800-ƒÊ F capacitor charged to 1.20�~104 Volts is converted to thermal energy by discharging the capacitor through a heating element submerged in 711.6 g of water in an insulating cup. What is the increase in temperature of the water if the heat capacity of the heater can be ignored? Assume that all of the energy stored on the capacitor goes into heating the water.
To find the increase in temperature of the water, we need to calculate the amount of thermal energy transferred from the capacitor to the water.
The thermal energy transferred can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the thermal energy transferred (in Joules),
m is the mass of the water (in grams),
c is the specific heat capacity of water (in J/g·°C), and
ΔT is the change in temperature (in °C).
First, let's calculate the mass of water in kilograms:
m = 711.6 g = 0.7116 kg
The specific heat capacity of water is approximately 4.18 J/g·°C.
With the formula and known values, we can calculate the change in temperature:
Q = m * c * ΔT
ΔT = Q / (m * c)
However, we still need to calculate the thermal energy transferred, Q.
The stored electrical energy of a capacitor can be calculated using the formula:
E = 0.5 * C * V^2
Where:
E is the electrical energy stored (in Joules),
C is the capacitance (in Farads), and
V is the voltage (in Volts).
Given that the capacitance is not provided, we assume it's 800 μF, which should be converted to Farads (F):
C = 800 μF = 800 * 10^(-6) F = 0.0008 F
Now we can calculate the electrical energy stored on the capacitor:
E = 0.5 * C * V^2
E = 0.5 * 0.0008 * (1.20 * 10^4)^2
Simplifying:
E = 0.5 * 0.0008 * 1.44 * 10^8
E ≈ 57600 Joules
Now that we have the thermal energy transferred (Q) and the mass of the water (m), we can calculate the change in temperature:
ΔT = Q / (m * c)
ΔT = 57600 / (0.7116 * 4.18)
ΔT ≈ 201.96 °C
Therefore, the increase in temperature of the water is approximately 201.96 °C.