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**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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