Question

The heating element of a coffeemaker operates at 120 V and carries a current of 4.50 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.525 kg of water from room temperature (23.0°C) to the boiling point.

Answer 1

Answer:

Explanation:

Voltage, V = 120 V

Current, i = 4.5 A

mass of water, m = 0.525 kg

initial temperature of water, T1 = 23°C

Final temperature of water, T2 = 100 °C

specific heat of water, c = 4.18 x 1000 J/kg °c

let the time taken is t.

Heat given by the heater = heat gain by the water

V x i x t = m x c x (T2 - T1)

120 x 4.5 x t = 0.525 x 4.18 x 1000 x (100 - 23)

540 t = 47701.5

t = 88.34 s

Answer 2

Answer:

It will take 313.376 sec to raise temperature to boiling point

Explanation:

We have given that potential difference V = 120 Volt

Current i = 4.50 A

So resistance $R=\frac{V}{i}=\frac{120}{4.50}=26.666ohm$

Heat flow in resistor will be equal to $Q=i^2Rt$

It is given that this heat is used for boiling the water

Mass of the water = 0.525 kg = 525 gram

Specific heat of water 4.186 J/gram/°C

Initial temperature is given as 23°C

Boiling temperature of water = 100°C

So change in temperature = 100-23 = 77°C

Heat required to raise the temperature of water $Q=mc\Lambda T$

So $4.50^2\times 26.666\times t=525\times 4.186\times 77$

t = 313.376 sec

So it will take 313.376 sec to raise temperature to boiling point