- #1
rahndezvous
- 2
- 0
Hi this is my first post and I'm stumped on my last HW problem, so I'm open to suggestions or solutions.
1. A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 35°C. In an attempt to cool the liquid, which has a mass of 220 g, 112 g of ice at 0.0°C is added. At the time at which the temperature of the tea is 15°C, determine the mass of the remaining ice in the jar. Assume the specific heat capacity of the tea to be that of pure liquid water.
_______grams
2. specific heat of water 4186 J/Kg*C
Latent heat of fusion for water 3.33*10^5 J/Kg
Q=Cp*m*/\T
Q=m*Lf
m=Q/Lf
4186 Cp*.22 Kg*20 C=18418.4
18418.4/3.33*10^5 Lf = .5531
.5531*1000=553.1g
Why do I get more than i started with?
1. A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 35°C. In an attempt to cool the liquid, which has a mass of 220 g, 112 g of ice at 0.0°C is added. At the time at which the temperature of the tea is 15°C, determine the mass of the remaining ice in the jar. Assume the specific heat capacity of the tea to be that of pure liquid water.
_______grams
2. specific heat of water 4186 J/Kg*C
Latent heat of fusion for water 3.33*10^5 J/Kg
Q=Cp*m*/\T
Q=m*Lf
m=Q/Lf
4186 Cp*.22 Kg*20 C=18418.4
18418.4/3.33*10^5 Lf = .5531
.5531*1000=553.1g
Why do I get more than i started with?