Question

How much heat transfer occurs to the environment by an electrical power station that uses 1.25×10¹⁴ J of heat transfer into the engine with an efficiency of 42.0%? (b) What is the ratio of heat transfer to the environment to work output? (c) How much work is done?

a) 7.25×10¹³ J, 2.50, 5.25×10¹³ J
b) 8.00×10¹³ J, 2.75, 5.50×10¹³ J
c) 8.75×10¹³ J, 3.00, 5.75×10¹³ J
d) 9.50×10¹³ J, 3.25, 6.00×10¹³ J

Answer 1

The heat transfer to the environment is 7.25x10^13 J. The ratio of heat transfer to the environment to work output is 0.58. The work done is 5.25x10^13 J.

Step-by-step explanation:

Let's solve the given problem step by step:

(a) The heat transfer to the environment can be calculated by multiplying the heat transfer into the engine (1.25x10^14 J) by (1 - efficiency).

Heat transfer to the environment = 1.25x10^14 J x (1 - 0.42).

Heat transfer to the environment = 7.25x10^13 J.

(b) The ratio of heat transfer to the environment to work output can be calculated by dividing the heat transfer to the environment by the work output.

Ratio of heat transfer to the environment to work output = 7.25x10^13 J / 1.25x10^14 J.

Ratio of heat transfer to the environment to work output = 0.58.

(c) The work done can be calculated by multiplying the heat transfer into the engine (1.25x10^14 J) by the efficiency.

Work done = 1.25x10^14 J x 0.42.

Work done = 5.25x10^13 J.