Question

Heat transfer is required to raise the temperature of a 0.699 kg aluminum pot containing 2.77 kg of water from 32.5ºC to the boiling point and then boil away 0.695 kg of water. If the rate of heat transfer is 292 W, how long does this process take? Aluminum's specific heat is 0.215 kcal/kgºC; Water's specific heat is 1.00 kcal/kgºC; and heat of vaporization for water is 539 kcal/kg.

Answer 1

The time is 8192.10 sec.

Step-by-step explanation:

Given that,

Mass of aluminum = 0.699 kg

Mass of water = 2.77 kg

Temperature = 32.5°C

Power = 292 W

We need to calculate the energy by aluminum

Using formula of energy

`E_(a)=mc\Delta T`

Put the value into the formula

`E_(a)= 0.699*0.215*(100-32.5)`

`E_(a)=10.144\ kcal`

We need to calculate the energy by water

Using formula of energy

`E_(w)=m_(w)c\Delta T+m_(w)L_(w)`

Put the value into the formula

`E_(w)=2.77*1.00*(100-32.5)+0.695*539`

`E_(w)=561.58\ kcal`

We need to calculate the total energy

`E=E_(a)+E_(w)`

`E =10.144+561.58`

`E=571.724\ kcal`

`E=2392093.216\ J`

We need to calculate the time

Using formula of energy

`E=P* t`

`t=(E)/(P)`

Put the value into the formula

`t=(2392093.216)/(292)`

`t=8192.10\ sec`

Hence, The time is 8192.10 sec.