Question

8. Water of mass 1.5 kg is heated from 30 degrees Celsius to 80 degrees Celsius using an electric kettle which is rated 5A, 239 volt. calculate the time taken to reach the final temperature specific heat capacity of water is 4200.
9. An immersion heater rated 400w, 220 volt is used to heat a liquid of mass 0.5 kg. if the temperature of the new point increases uniformly at the rate of 2.5 degrees Celsius per second calculate the specific heat capacity of the liquid.
10. How long does it take a 750W heater operating at full written to raise the temperature of 1 kg of water from 40 degrees Celsius to 70 degrees Celsius .(specific heat capacity of water is 4200)

Answer 1

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.