You have two cups, one containing 100 g of ice at 0oC, the other containing 100 g of water at 80oC. You pour the hot water into the cup containing the ice. What do you wind up with?

A) 100 g of 0oC water and 100 g of 80oC water.

B) 50 g of 0oC ice and 50g of 0oC water.

C) 200 g of 0oC water.

D) 100 g of 0oC water.
I think the answer is D am I correct

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  1. Seems to me you ultimately need to find the total for both the mass of the H2O and the mean for temperature. You will initially find the temperature of the combination water decreasing and volume of the ice decreasing. Once out of the solid state, the liquid ice will increase in temperature the entire combination will approach the mean.

    Before reaching the mean of the temperature, you will find that the temperature close to the ice will be lower than that in the liquid state water further from the ice. Close to the ice, you should expect to find 0 degrees C. What will be interesting for you to explore is how long at it will take for the transition to that mean to take place.

    You have two identical substances that have similar properties except that one does not generally exist in the solid form above its freezing point. Each gram of pure 80 degree water (each ml) will need to dissipate 1 calorie of energy per degree to reach the 40 degree mean. Each gram of pure water ice will need to absorb 1 calorie per degree to for it to reach 40 degrees. Since it is not an instantaneous heat exchange, both the different parts of the combination will have different physical and thermal properties until they reach that mean.

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