# If you drop a 200 gram piece of metal with a temperature of 110 degrees Celsius into 1000 grams of water at 25° Celsius, what best describes what would occur? The water will quickly reach the boiling point. The water's temperature will stay the same, b

## If you drop a 200 gram piece of metal with a temperature of 110 degrees Celsius into 1000 grams of water at 25° Celsius, what best describes what would occur? The water will quickly reach the boiling point. The water's temperature will stay the same, but the metal will cool down. The water's temperature will increase, and the temperature of the metal will stay constant. The water and the metal's temperatures will reach the same temperature.

Jul 23, 2018

d) The water and the metal's temperatures will reach the same temperature.

#### Explanation:

Heat exchange occurs between two objects in an insulated system as long as differences exist between temperatures of the two objects. The mixture would eventually see no heat exchange between the metal and the water. Thus the temperature of neither object will stay unchanged.

It is highly unlikely for the water to boil in response to the addition of this piece of metal ${10}^{\text{o" "C}}$ above the $\text{b.p.}$ of water under standard conditions. Assuming that the system is properly insulated from the surroundings, the amount of heat the water absorbs as its temperature rise to the final temperature, $t$, would equal to that the metal releases as it cools. Thus

$c \left(\text{metal") * 200 color(white)(l) "g" * (110 - t) = "Q"("metal}\right)$
$c \left(\text{water") * 1000 color(white)(l) "g" * (t - 25) = "Q"("water}\right)$
"Q"("metal") = "Q"("water")

The heat capacity of the metal would be higher than

c("metal") * 200 color(white)(l) "g" * (110 - 100) >= c("water") * 1000 color(white)(l) "g" * (100 - 25)
$c \left(\text{metal") >= 37.5 * c("water}\right)$

for the metal to supply sufficient energy for the water to boil. It is unheard of for the specific heat of metals to exceed that of water by so many folds. The assumption doesn't hold, and therefore the temperature of the final mixture would be lower than the $\text{b.p.}$ of water.