Respuesta :

syed514
 Speicific heat of water is 4.2j/gm*C 

Thus 4.2 Joules of energy is required to heat 1 gm water to 1*C. 
or 4.2 x 30 Joules enery is required to heat 1x30 gm water by 1*C 
or 4.2 x 30 x 80 Joules enery is required to heat 1 x 30 gm water by (100-20)*C( or required to be removed from water to cool it) 

Answer:

[tex]-77.74~KJ[/tex]

Explanation:

In this case we have two processes. First, we have to convert the gas water into liquid water at 100 ºC and then we have to go from 100 ºC to 20 ºC  in the liquid state.

For the first part, we have to use the enthalpy of vaporization ([tex]2257\frac{J}{g}[/tex]), so:

[tex]30~g~H_2O~x~2257\frac{J}{g}[/tex]

[tex]67710~J[/tex]

In this case we go from gas to liquid, so energy is released, therefore:

[tex]-67710~J[/tex]

In the second part we have to calculate the energy in the change from 100 ºC to 20 ºC, so:

[tex]Q=mCp[/tex]ΔT

[tex]Q=30g~4.18\frac{J}{g*^{\circ}C}*(20-100)^{\circ}C[/tex]

[tex]-10032~J[/tex]

So, in total we will have:

[tex](-10032~J)~+~(-67710~J)[/tex]

[tex]-77742~J[/tex] or [tex]-77.74~KJ[/tex]

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