A 1250.0g block of silver at 1oC is placed in 2.6g of water vapor at 102oC. What is the equilibrium temperature of the two substances? Specific Heat of Ag is 0.235J/goC Please answer with work shown

Explanation:

Let final temperature be T .

vapor is at 102⁰C

loss of heat by vapor in turning into water at 100⁰C

= 2.6 x 2 x 1.996 + 2.6 x 2260 = 5886.37 J

loss of heat to lower temperature to T

2.6 x 4.186 x ( 100 - T )

1088.36 - 10.88 T

Total heat loss = 5886.37 + 1088.36 - 10.88 T

= 6974.73 - 10.88 T

heat gain by silver to gain temperature from 1⁰C to T⁰C

= 1250 x ( T - 1 ) x .235 = 293.75 T - 293.75

heat gain = heat loss

293.75 T - 293.75  = 6974.73 - 10.88 T

304.63 T = 7268.48

T = 23.86°C .

1. they vary with a higher number because boiling is way hotter than melting