Question

Assume that an average city uses about 8.5×1013J of energy in a day.

A fission-based nuclear bomb (such as those used in World War II) uses the same nuclear reactions and an amount of fuel similar to what you calculated in Part A. We will assume for our approximate calculation that matter to energy conversion is at 0.01%, as in Part A. However, by allowing an uncontrolled chain reaction to occur, all of the bomb's energy can be released in about 1.0×10−6s.
What is the average power Pavg of such a nuclear bomb?
Express your answer in watts to two significant figures.

Answer 1

the average power of such nuclear bomb is 8.5 × 10¹⁹Watt

Step-by-step explanation:

Given that,

Energy release in the nuclear reaction,E is 8.5×10^13J

time, t = 1.0×10−6s.

The average power of the nuclear bomb is ,

Expression for power is given as

Average Power = E / t

Average Power =

`(8.5*10^1^3J)/(1.0* 10^-^6s.)`

Average Power = 8.5 × 10¹⁹Watt

Hence, the average power of such nuclear bomb is 8.5 × 10¹⁹Watt

Answer 2

Pavg = (8.5 × 10²³ W)

Step-by-step explanation:

The city uses (8.5×10¹³)J of energy in a day,

And this represents only 0.01% of the total nuclear energy released.

All of the energy released in the nuclear fission is (1.0×10⁻⁶) s.

Let the total energy produced be x

0.01% of x = 8.5×10¹³ J

x = (8.5×10¹³) ÷ 0.01% = (8.5×10¹⁷) J

Average power produced = (energy produced)/Time

Pavg = (8.5×10¹⁷)/(1.0×10⁻⁶)

Pavg = (8.5 × 10²³ W)

Hope this Helps!!!