Question

if 2.93*10^4 joules are added to 10.8 gram of water at 32 0C, what is the final temperatuer of the water sample?

Answer 1

Given : Heat added = 2.93 E 4 J , Mass of water = 10.8 g , Intial temperature of water = 32.0 deg C ,

We use specific heat of water and vapor from reference table.

C for water = 4.184 J /g deg C , C for vapor = 2.09 J / deg C

final temperature of water = unknown

We use following equation

q = c m delta T

c = specific heat , m = mass in g , Delta T = Tf- Ti

Lets plug all the given values to get Tf ( final temperature )

Before calculating the value of final temperature lets find the heat needed to reach the 100 deg c and heat needed to evaporate all water.

q 1 = 10.8 g * 4.184 J /g deg C * ( 100-32.0 )

= 3072.72 J

Heat needed to evaporate the water

q 2 = Delta H * m = 333.3 J / g * 10.8 g = 3599.64 J

Now lets us add the q1 and q2

= 3072.72 J + 3599.64 J = 6672.37

So this heat is less than given heat and so ther eare three transition s.

But given heat is given more than this. so there are three transitions

1st transition q1 from 32.0 deg C to 100 deg C

2.93 E 4 J =  3072.72 J + 3599.64 J + 10.8 g * 2.09 J / g deg C * (Tf - 100 deg C )

22627.63 J = 2.09 J / g deg C * (Tf - 100 deg C )

10826.62 deg C = Tf - 100 deg C

10926.62 deg C = Tf

Final temperature = 10926.62 deg C

So final temperature of water = 10926.62 deg C