Question

A 2400-W oven is connected to a 240-V source.

1. What is the resistance of the oven?

Express your answer to two significant figures and include the appropriate units.

2. How long will it take to bring 110 mL of 25 ∘C water to 100∘C assuming 65 % efficiency? The specific heat capacity of water is 4186 J/(kg⋅∘C).

Express your answer using two significant figures.

3. How much will this cost at 12 cents/kWh?

Express your answer using two significant figures.

Answer 1

Solution:

Let us go to the basics first.

A 2400-W oven is connected to a 240-V source.

1. What is the resistance of the oven?

Answer: We know that Power P is given by following relation:

P = V² / R

=>2400 = 240² / R

=>R = 24 Ohms (Answer)

2. How long will it take to bring 110 mL of 25 ∘C water to 100∘C assuming 65 % efficiency? The specific heat capacity of water is 4186 J/(kg⋅∘C).

Answer: We know that heat required Q is given by:

Q = mcdt

[where, m = mass of water = 110 mL = 0.11 kg; c = 4186 J/(kg⋅∘C); dt = 100°C - 25°C = 75°C]

=>Q = 0.11*4186*75 = 34534.5 J

Since oven has efficiency = 65%

So, effective power = 2400 * 0.65 W = 1560 W

Thus, time required = Q / Effective power = 34534.5 J / 1560 W = 22 seconds (Answer)

3. How much will this cost at 12 cents/kWh?

Answer: Electricity consumed = 2400 W * 22 seconds = 2.4KW*0.006h = 0.0144 KWh

Thus, cost = 0.0144 KWh*12 cents/kWh = 0.1728 cents = 0.17 cents (Answer)

Thanks!!!