Suppose that a commercial espresso machine in a coffee shop turns 1.50 kg of water at 22.0? c into steam at 100? c. if c=4187j/(kg? ? c) and hv=2,258kj/kg, how much heat q is absorbed by the water from the heating resistor inside the machine? assume that this is a closed and isolated system. express your answer in joules to three significant figures. q = j

Total heat required will be 555.756 KJ

Explanation:

We have given mass of the water m = 1.50 kg

Temperature is changes from to steam

Specific heat is given as c = 4718 J/kg

So amount of heat for changing the temperature from to

Heat of vaporization is given

So the amount of heat required to vaporize water

So total heat

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