How much water, initially at 20C, could be boiled by this quantity of heat

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In summary: So when you want the amount of water that can be boiled with a given amount of energy, you divide the energy by the heat of vaporisation. So the final answer in kg would be 2.857 kg. In summary, if an average city with a population of less than one million uses 8.5*10^13J of energy in a day, and 35% of that energy is released as heat in a purely fission-based nuclear explosion, it would be enough to boil approximately 2.857 kg of water, initially at 20C, using the heat of vaporization of 2.26*10^6 J/kg and the specific heat of 4186 J/kg*C.
  • #1
Nuclear Power and Nuclear Weapons
Assume that an average city (with a little less than one million people) uses about 8.5*10^13J of energy in a day.
In a purely fission-based nuclear explosion, about 35% of the energy is released in the form of heat. How much water, initially at 20C, could be boiled by this quantity of heat? Use 2.26 *10^6J/kg for the heat of vaporization of water and 4.18J/(g*C) for the specific heat of water.

Hint. Calculate the energy needed to boil 1kg of water
Find the energy Q needed to boil 1.0kg of water. Be careful with units, because grams and kilograms are used for mass in different constants.
For this hint I tried these answers... all of which were wrong:
1) (1kg)*(1000g/kg)*(4.18J/g*C)*(80C)=3.34×10^5
2) 284648
3) (3.34×10^5)/(.35)=954285.7143
Hint. Heat needed to change the temperature of a substance
The heat (energy transfer) needed to change the temperature of a substance, without any phase change, is given by Q = m*c*Delta T, where m is the mass, c is the specific heat, and DeltaT is the change in temperature of the substance.

Hint 2. Dealing with phase change
To find the energy required to boil the water, simply find the energy needed to raise its temperature to 100C, and then add the energy needed to change state from liquid to gas.

Any help is appreciated... :!)
 
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  • #2
Lets start with the hint you were given. You have 1 kg of water at 20 C. Now to boil the water, you need to first heat it to 100 C. Then you have to give it 2.26 * 10^6 Joules for the phase change from liquid to vapor. This is the heat of vaporisation. Can you take it from here?
 
  • #3
Following hint 1:
Q=m*C*deltaT ... Q should end up in Joules
Q=(1kg)*(4186J/kg*C)*(80C)=334880 Joules (but this is not the wrong answer for Q)
so then if I take that answer and multiply it by (2.26*10^6 J/kg)= 7.57*10^11 J/kg but Q is supposed to be in Joules.
 
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  • #4
I think that for the overall problem I have to find the Q in joules and multiply that by 2.26*10^6 J/kg to get some number of kilograms as the answer... but first (for the hint) i need to find Q energy in joules and that's what I am having trouble finding...

also I'm not sure where the 35% comes into the question
 
  • #5
334880 J is the amount of energy to raise it to 100 C. Now you are giving it more energy to change its phase. So you're supposed to add 2.26 x 10^6 J, not multiply by it.
 
  • #6
If you do that, you will obtain the total amount of energy to take 1 kg of water at 20 C to water vapor at 100 C. Now, how much energy is released by the nuclear explosion in the form of heat?
 
  • #7
35% of energy is released as heat

(334880 J) + (2.26 x 10^6 J/kg) = 2594880 joules
*but specific heat is in units of (2.26 x 10^6 J/kg) not just (2.26 x 10^6 J)
How can you add J + J/kg ?

2594880 J = 0.35 (Total Energy Released)
Total energy released= 7413942.857 J

The amount of water in kg that could be heated from this energy is...
then should i divide total energy release by energy to raise 1kg of water
 
  • #8
mtbgymnast said:
(334880 J) + (2.26 x 10^6 J/kg) = 2594880 joules
*but specific heat is in units of (2.26 x 10^6 J/kg) not just (2.26 x 10^6 J)
How can you add J + J/kg ?

It's not J/kg. The unit of specific heat is J/(kg*C). To get the heat needed to raise temp from 20 to 100 you had to multiply by the mass ( 1kg) and the change in temp (80 C), so the Kg and C cancel out, and your'e left with a quantity whose unit is J as it should be.

mtbgymnast said:
The amount of water in kg that could be heated from this energy is...
then should i divide total energy release by energy to raise 1kg of water

yes.
 
  • #9
oops I'm sorry
i meant heat of vaporization of water not specific heat
heat of vaporization of water= 2.26*10^6 j/kg
specific heat of water=4.186 J/g*C =4186J/kgC

So far here is what I have:
(1kg)*(4186J/(kg*C))*(80C)=334880 J
334880 J + 2.26*10^6 J/kg = 2594880 J
2594880 J = 0.35 (Total Energy Released)
Total energy released= 7413942.857 J
7413942.857 J / 2594880 J/kg = 2.857 J
but I need it in kg
 
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  • #10
It's the same explanation for heat of vaporisation. When you want the energy needed to vaporise 1 kg of water, you multiply the heat of vaporisation by 1 kg, and you get an answer in joules.
 

1. How does the initial temperature affect the amount of water that can be boiled?

The initial temperature of the water does not have a direct effect on the amount of water that can be boiled by a certain quantity of heat. However, the initial temperature does affect the time it takes for the water to reach its boiling point.

2. How does the quantity of heat impact the amount of water that can be boiled?

The quantity of heat has a direct correlation with the amount of water that can be boiled. The more heat that is added to the water, the higher the temperature of the water will rise and the more water will be able to reach its boiling point.

3. Is there a limit to the amount of water that can be boiled by a specific quantity of heat?

Yes, there is a limit to the amount of water that can be boiled by a specific quantity of heat. This limit is known as the heat capacity of water, which is approximately 4.186 joules per gram per degree Celsius. Once this limit is reached, the water will not be able to absorb any more heat and will remain at its boiling point.

4. Can the type of container used affect the amount of water that can be boiled by a specific quantity of heat?

Yes, the type of container used can affect the amount of water that can be boiled by a specific quantity of heat. Different materials have different thermal conductivities, which can affect the rate at which heat is transferred to the water. For example, a metal container will transfer heat more efficiently than a plastic container, resulting in a higher amount of water being boiled.

5. How can the amount of water that can be boiled be calculated?

The amount of water that can be boiled can be calculated by using the specific heat capacity of water and the quantity of heat added. The equation for this is Q = mcΔT, where Q is the quantity of heat, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature. By rearranging this equation, we can solve for the mass of water that can be boiled with a given quantity of heat.

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