Physics: 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.

not rly, tho even if you told me, i would forget in half a second lolExplanation:...

Water is boiled at 300 kPa pressure in a pressure - id3005312341, fattypickeltoefungus

owen74

20.01.2022

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

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

$h_{2}=561.47+0.5\times2163.8$

$h_{2}=1643.37\ kJ/kg$

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

Using formula of rate of energy

$Q=\dfrac{m(h_{2}-h_{1})}{t}$

Put the value into the formula

$Q=\dfrac{3\times(1643.37-561.47)}{30\times60}$

$Q=1.80\ kW$

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

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