Question

Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it is observed that half of the water in the cooker evaporated in 30 minutes. Assume negligible heat loss from the cooker.

Answer 1

The average rate of energy transfer to the cooker is 1.80 kW.

Step-by-step explanation:

Given that,

Pressure of boiled water = 300 kPa

Mass of water = 3 kg

Time = 30 min

Dryness friction of water = 0.5

Suppose, what is the average rate of energy transfer to the cooker?

We know that,

The specific enthalpy of evaporate at 300 kPa pressure

`h_(f)=561.47\ kJ/kg`

`h_(fg)=2163.8\ kJ/kg`

We need to calculate the enthalpy of water at initial state

`h_(1)=h_(f)`

`h_(1)=561.47\ kJ/kg`

We need to calculate the enthalpy of water at final state

Using formula of enthalpy

`h_(2)=h_(f)+xh_(fg)`

Put the value into the formula

`h_(2)=561.47+0.5*2163.8`

`h_(2)=1643.37\ kJ/kg`

We need to calculate the rate of energy transfer to the cooker

Using formula of rate of energy

`Q=(m(h_(2)-h_(1)))/(t)`

Put the value into the formula

`Q=(3*(1643.37-561.47))/(30*60)`

`Q=1.80\ kW`

Hence, The average rate of energy transfer to the cooker is 1.80 kW.