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jeyben [28]
3 years ago
9

How much heat is required to convert 2.55g of water at 28 degrees c to steam?

Physics
1 answer:
pantera1 [17]3 years ago
4 0
There are two different processes here:
1) we must add heat in order to bring the temperature of the water from 28^{\circ}C to 100^{\circ}C (the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates

1) The heat needed for process 1) is
Q_1=m C_s \Delta T
where 
m=2.55 g is the water mass
C_s = 4.18 g/J^{\circ}C is the water specific heat
\Delta T=100^{\circ}C-28^{\circ}C=72^{\circ}C is the variation of temperature of the water
If we plug the numbers into the equation, we find
Q_1 = (2.55 g)(4.18 J/g^{\circ}C)(72^{\circ}C)=767.4 J

2) The heat needed for process 2) is
Q=m L_e
where 
m=2.55 g is the water mass
L_e = 2264.7 J/g is the latent heat of evaporation of water
If we plug the numbers into the equation, we find
Q_2=(2.55 g)(2264.7 J/g)=5775.0 J

So, the total heat needed for the whole process is
Q=Q_1+Q_2=767.4 J+5775.0 J=6542.4 J
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