The stored electrical energy of a 350-μ F capacitor charged to 1.20×104 Volts is converted to thermal energy by discharging the capacitor through a heating element submerged in 548.0 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 can use the equation:
Q = mcΔT
Where:
Q is the heat energy transferred to the water
m is the mass of the water
c is the specific heat capacity of water
ΔT is the increase in temperature
Given:
m = 548.0 g
c is the specific heat capacity of water = 4.18 J/g°C
Q is the electrical energy stored in the capacitor
First, let's calculate Q, the electrical energy stored in the capacitor:
Q = 0.5CV²
Where:
C is the capacitance of the capacitor
V is the voltage across the capacitor
Given:
C = 350 μF = 350 × 10⁻⁶ F
V = 1.20 × 10⁴ Volts
Plugging in the values:
Q = 0.5 × (350 × 10⁻⁶ F) × (1.20 × 10⁴ Volts)
Q = 0.5 × (0.00035 F) × (12000 V)
Q = 2.1 Joules
Now, let's substitute the calculated value of Q into the heat equation to find the increase in temperature:
2.1J = (548.0 g) × (4.18 J/g°C) × ΔT
Divide both sides of the equation by (548.0 g) × (4.18 J/g°C):
2.1J / [(548.0 g) × (4.18 J/g°C)] = ΔT
Calculate the value:
ΔT = 2.1J / ( 548.0 g × 4.18 J/g°C)
≈ 0.001 J / g°C
Therefore, the increase in temperature of the water is approximately 0.001°C.