Question

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

Answer 1

Answer:

Total heat required will be 555.756 KJ

Explanation:

We have given mass of the water m = 1.50 kg

Temperature is changes from $22^{\circ}C$ to $100^{\circ}C$ steam

Specific heat is given as c = 4718 J/kg

So amount of heat for changing the temperature from $22^{\circ}C$ to $100^{\circ}C$

$Q_1=mc(T_2-T_1)=1.5\times 4718\times (100-22)=552006J$

Heat of vaporization is given $H_V=2.25kj/kg$

So the amount of heat required to vaporize water

$Q_2=mH_V=1.5\times 2.5=3.75KJ$

So total heat $Q=Q_1+Q_2=552.006+3.75=555.756KJ$