course
Subject: Phy 122
Dr. Smith,
I finally got back on track on getting things done and when
I was working on Ch. 14 #13 I came across a problem. Here
is the question:
A hot iron horseshoe (mass=0.40kg) just forged is dropped
into 1.35L of water in a 0.30kg iron pot initially at
20degrees C. If the final equilibrium temp. is 25 degrees,
estimate the intitial temp. of the hot horseshoe.
Here's as far as I got:
1L =1x10^-3m^3 1.35L(1x10^-3m^3) = 0.00135kg water
The density of water is 1000 kg / m^3, so the mass of 1.35 liters comes out to 1.35 kg, not .00135 kg.
25 degrees C -20= 5 degrees
Q of iron pot = 0.30kg(450)(5)=657J
Q of water= 0.00135kg(4186)(5)=28.2555J
657+28.2555J = 703.2555J
Recalculate this total based on the correct mass of the water. Otherwise the procedure is fine so far.
The result will be equal to the energy lost by the horseshoe, which is equal to the mass of the horseshoe, multiplied by its change in temperature, multiplied by its specific heat. You know its specific heat and mass, so you can find the change in its temperature.
Let me know if this doesn? help.
From there I do not know where to go to get the answer
assuming that I am even on the right track. Please guide
me!!
Thanks as always,
*****