Question

A 100 g piece of copper, initially at 95,0 C ,
is dropped into 200 g of water. If initial temperature of
water is 1 5 C, what is final temperature of the system, if the heat lost to surounding is ignored

please, step by step solution.

Answer 1

a) The colder water will warm up (heat energy "flows" in to it). The warmer metal will cool down (heat energy "flows" out of it).
b) The whole mixture will wind up at the SAME temperature. This is very, very important.
c) The energy which "flowed" out (of the warmer water) equals the energy which "flowed" in (to the colder water)

Solution Key Number One: We start by calling the final, ending temperature 'x.' Keep in mind that BOTH the iron and the water will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using IS NOT the ?t, but the FINAL temperature. This is what we are solving for.

The warmer iron goes down from to 95.0 to x, so this means its ?t equals 85.0 minus x. The colder water goes up in temperature, so its ?t equals x minus 15.0.

That last paragraph may be a bit confusing, so let's compare it to a number line:

To compute the absolute distance, it's the larger value minus the smaller value, so 95.0 to x is 95.0 minus x and the distance from x to 15.0 is x minus 15.0

Solution Key Number Two: the energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

q_lost = q_gain

The water specific heat will remain at 4.184 and copper is 0.386

So, by substitution, we then have:

(100.0) (95.0 - x)(0.386) = (200.0) (x - 15.0) (4.184)

Solve for x

then

176.29=6.41 x

The answer is

27.5