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

[mtbgymnast]

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... :!)

 

[dx]

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?

 

[mtbgymnast]

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.

 

[mtbgymnast]

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

 

[dx]

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.

 

[dx]

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?

 

[mtbgymnast]

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

 

[dx]

(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.

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.

 

[mtbgymnast]

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

 

[dx]

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.