8. First, calculate the amount of heat energy required to heat the water:
Q = m * c * ΔT
Q = 1.5 kg * 4200 J/kg°C * (80°C - 30°C)
Q = 1.5 kg * 4200 J/kg°C * 50°C
Q = 315,000 J
Next, calculate the power of the electric kettle:
P = V * I
P = 239 V * 5 A
P = 1195 W
Then, calculate the time taken to heat the water:
t = Q / P
t = 315,000 J / 1195 W
t ≈ 263 seconds
Therefore, it will take approximately 263 seconds (or 4 minutes and 23 seconds) for the water to reach the final temperature.
9. Power (P) = 400W, Voltage (V) = 220V, Mass (m) = 0.5kg, Rate of temperature increase (dT/dt) = 2.5°C/s
Q = m * c * ΔT / t
P = Q / t
From the given information:
P = 400W = Q / t
Q = 400 J/s * t
Q = m * c * ΔT
400 J/s * t = 0.5kg * c * 2.5°C/s
400 J/s * t = 1.25kg * c
Specific heat capacity of the liquid:
c = (400 J/s * t) / 1.25kg
c = 320 J/kg°C
Therefore, the specific heat capacity of the liquid is 320 J/kg°C.
10. First, calculate the amount of heat energy required to heat the water:
Q = m * c * ΔT
Q = 1 kg * 4200 J/kg°C * (70°C - 40°C)
Q = 1 kg * 4200 J/kg°C * 30°C
Q = 126,000 J
Then, calculate the power of the heater:
P = 750W
Next, calculate the time taken to heat the water:
t = Q / P
t = 126,000 J / 750W
t = 168 seconds
Therefore, it will take 168 seconds (or 2 minutes and 48 seconds) for the water to reach the final temperature.