How much water would be required to absorb this heat?

  • Thread starter Intrusionv2
  • Start date
  • Tags
    Heat Water
In summary, the United States generates about 5.7 x 10^16 J of electric energy a day, which is equivalent to work. However, with an average efficiency of 0.32, only a fraction of this energy is actually converted into work, resulting in a significant amount of heat being dumped into the environment each day. To absorb this heat without increasing water temperature by more than 2.7° C, approximately 1.07e+13 kg of water would be required. This is due to the fact that most of the generated power is eventually expelled as waste heat, rather than being used for productive purposes.
  • #1
Intrusionv2
31
0

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
 
Physics news on Phys.org
  • #2


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.
 
  • #3


Intrusionv2 said:
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
 
  • #4


Andrew Mason said:
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!
 
  • #5
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.
 

1. How do you calculate the amount of water needed to absorb a certain amount of heat?

The amount of water needed to absorb a certain amount of heat can be calculated using the specific heat capacity of water, which is 4.186 joules per gram per degree Celsius. This means that 4.186 joules of energy is required to raise the temperature of 1 gram of water by 1 degree Celsius. By knowing the specific heat capacity and the amount of heat to be absorbed, the amount of water needed can be calculated using the formula Q = m x c x ΔT, where Q is the heat energy, m is the mass of water, c is the specific heat capacity, and ΔT is the change in temperature.

2. How does the initial temperature of the water affect the amount of water needed to absorb heat?

The initial temperature of the water does not affect the amount of water needed to absorb heat. The specific heat capacity of water is constant and only depends on the mass and temperature change. Therefore, the same amount of water will be needed regardless of the initial temperature.

3. Can any type of water be used to absorb heat?

Yes, any type of water can be used to absorb heat. However, the specific heat capacity of water may vary slightly depending on its impurities and salinity. Pure water has a specific heat capacity of 4.186 joules per gram per degree Celsius, while seawater has a slightly higher specific heat capacity of 3.9 joules per gram per degree Celsius.

4. Is there a limit to how much heat water can absorb?

There is no limit to how much heat water can absorb. However, the temperature of the water can only increase to a certain point before it reaches its boiling point and turns into steam. At this point, the water can continue to absorb heat, but its temperature will remain constant until all the water has turned into steam.

5. How does the amount of water needed to absorb heat change with the duration of the heat source?

The amount of water needed to absorb heat does not change with the duration of the heat source. The specific heat capacity of water remains constant regardless of how long the heat source is applied. However, the temperature of the water will continue to increase the longer the heat source is applied, so more heat energy will need to be absorbed to maintain the desired temperature. This means that the amount of water needed may change if the desired temperature is different for different durations of the heat source.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
882
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
398
  • Introductory Physics Homework Help
Replies
1
Views
917
  • Introductory Physics Homework Help
Replies
1
Views
882
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
22
Views
2K
Back
Top