Temperature & Heat

[Kris1120]

1. The problem statement, all variables and given/known data

A thermos contains 155 cm3 at 87 degrees C. To cool the coffee, you drop two 12.2-g ice cubes into the thermos. The ice cubes are initially at 0 degrees C and melt completely. What is the final temperature of the coffee in degrees Celsius? Treat the coffee as if it were water

2. Relevant equations

(m*Lf)ice= (c*m*Delta T)water

3. The attempt at a solution

(.0244)(33.5x10^4)= (4186)(155)(Tf-87.0)

T f= 86.987 degrees C

[Hootenanny]

1. The problem statement, all variables and given/known data

A thermos contains 155 cm3 at 87 degrees C. To cool the coffee, you drop two 12.2-g ice cubes into the thermos. The ice cubes are initially at 0 degrees C and melt completely. What is the final temperature of the coffee in degrees Celsius? Treat the coffee as if it were water

2. Relevant equations

(m*Lf)ice= (c*m*Delta T)water

3. The attempt at a solution

(.0244)(33.5x10^4)= (4186)(155)(Tf-87.0)

T f= 86.987 degrees C
What happens to the water that results from the ice melting?

[Kris1120]

When the ice melts, it cools the coffee so the overall temperature decreases.

[Hootenanny]

When the ice melts, it cools the coffee so the overall temperature decreases.
Correct. Does the temperature of the melted ice stay at 0oC or does it change?

[Kris1120]

The ice gains heat and the coffee loses heat

[Hootenanny]

The ice gains heat and the coffee loses heat
Correct, but you haven't taken this into account in your calculations have you? You need to take into account the temperature change of the ice-water after it has melted.

[Softer1]

i think after ice melt , temperature of water (for Ice melt) will up and water in thermos will drop

SO

(m*Lf)ice + (m*c*(Tf-0))ice = (c*m*Delta T)water

(.0244)(33.5x10^4)+(.0244)(4186)(Tf-0)= (4186)(155)(87-Tf)