How much water would be required to absorb this heat?

[Intrusionv2]

Homework Statement

The United States generates about 5.7 x 10^16 J of electric energy a day. This energy is equivalent to work, since it can be converted into work with almost 100% efficiency by an electric motor.

(a) If this energy is generated by power plants with an average efficiency of 0.32, how much heat is dumped into the environment each day?

Answer --> 1.21e+17 J

(b) How much water would be required to absorb this heat if the water temperature is not to increase more than 2.7° C?

Answer --> 1.07e+13 kg.

The Attempt at a Solution

I don't see how it is possible to pump out more heat into the environment than work generated, especially at 0.32 efficiency. Can someone help me with this problem please? Thanks

 

[rl.bhat]

a) Efficiency = work output/work input.
In the problem, work output and efficiency is given. Find work input.
heat is dumped into the environment each day = work input - work output.
b) Q = m*c*change in temperature.

 

[Andrew Mason]

I don't see how it is possible to pump out more heat into the environment than work generated, especially at 0.32 efficiency. Can someone help me with this problem please? Thanks

Following on what rl.bhat has said, the work is generated from heat engines. The heat engines only convert a fraction (.32) of the heat flow into work. The rest is expelled as waste heat.

AM

 

[gneill]

Following on what rl.bhat has said, the work is generated from heat engines. The heat engines only convert a fraction (.32) of the heat flow into work. The rest is expelled as waste heat.

AM

One might argue that virtually *all* of the generated power is eventually expelled as waste heat; mostly it is used to heat things, illuminate things, and do repetitive tasks in a friction-filled environment. Elevators go down as well as up!

 

[rwisz]

for your help

I would like to clarify that the amount of heat generated by power plants is not necessarily equal to the amount of work generated. In fact, according to the first law of thermodynamics, energy cannot be created or destroyed, it can only be converted from one form to another. In this case, the electric energy generated by power plants is converted from other sources such as coal, nuclear energy, or renewable sources.

To answer the question, we can use the formula Q = mcΔT, where Q is the amount of heat absorbed, m is the mass of water, c is the specific heat capacity of water (4.186 J/g·K), and ΔT is the change in temperature (2.7°C).

(a) Using the given information, we can calculate the amount of heat dumped into the environment each day as follows:

Q = (5.7 x 10^16 J) / 0.32 = 1.78125 x 10^17 J

(b) To find the amount of water required to absorb this heat without increasing the temperature by more than 2.7°C, we rearrange the formula as follows:

m = Q / (c ΔT)

m = (1.78125 x 10^17 J) / (4.186 J/g·K x 2.7°C) = 1.07 x 10^13 g = 1.07 x 10^10 kg

Therefore, approximately 10 billion kilograms of water would be required to absorb the heat generated by power plants in the United States each day without increasing the temperature by more than 2.7°C. This amount may seem large, but it is important to note that this is a theoretical calculation and does not take into account factors such as heat loss during absorption or the use of cooling systems in power plants.