Question

Which of the following processes will result in the lowest final temperature of the metal–water mixture at when thermal equilibrium is reached? The specific heat capacity of cobalt is 0.421 J/(g·°C). The specific heat capacity of water is 4.184 J/(g·°C).

a. the addition of 100 g of cobalt at 95°C to 80 mL of water at 25°C in an insulated container

b. the addition of 100 g of cobalt at 95°C to 100 mL of water at 25°C in an insulated container

c. the addition of 100 g of cobalt at 95°C to 40 mL of water at 25°C in an insulated container

d. the addition of 100 g of cobalt at 95°C to 20 mL of water at 25°C in an insulated container e. the addition of 100 g of cobalt at 95°C to 60 mL of water at 25°C in an insulated container

Answer 1

Just forget those numbers for now.

We know that the metal (cobalt) is hot.

Water can act as a coolant.

So what if the amount of water is increased? It can cool better and results in the lowest final temperature.

So the highest volume (or mass) of water will result in the lowest equilibrium temperature.

(Note that this is only applicable in the case of a constant mass and temperature of the metal (cobalt), otherwise we need to try the equations).

So the correct answer will be:

(b) the addition of 100 g of cobalt at 95°C to 100 mL of water at 25°C in an insulated container.