Hot ball in cold water

Classical Mechanics Level 1

A metal ball with temperature $80 ^\circ\text{C}$ is dropped in water with temperature $30 ^\circ\text{C}$ . The metal ball and water exchange heat with each other, and after a while both their temperatures become constant.

Assuming that no heat has been lost to or gained from the surroundings and the container, if the water temperature has been raised by $10 ^\circ\text{C}$ , what is the decrease in the temperature of the ball?

Equal to

10 ^\circ\text{C}

Less than

10 ^\circ\text{C}

Greater than

10 ^\circ\text{C}

Insufficient data; mass and material of the ball must be given

1 solution

Marta Reece
Jun 20, 2017

Relevant wiki: Equilibrium

If the temperature of the water was increased by $10^\circ$ from $30^\circ$ , it is now $40^\circ$ .

Once the system is stabilized, the temperature of the ball must be also $40^\circ$ .

This is $40^\circ$ less than the original $80^\circ$ .

The decrease in the temperature of the ball is, therefore, larger than $10^\circ$ .

You can arrive at the same answer by noting that in the illustration the water appearals to possess greater mass than the ball, so to increase the temperature of the water by 10 degrees the ball must lose more than 10 degrees of heat. Of couse, the problem only states that the material of the ball is metal, so it could presumably be sodium or another metal known to be extremely reactive in water, or it could be a radioactive isotope that keep the metal and the water both at 80. So technically, we've gotta give it to those who answer that we cannot know without more information.

Jonathan Fitch - 3 years, 11 months ago

Hot ball in cold water

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