Question 1085178
A) If {{{m}}} is the mass, in grams, of each component (water and iron,
and {{{x}}} is the final temperature of both the water and the iron, in degrees Celsius,
the heat gained by the water, in calories, is
{{{m*1*(x-20)}}} ,
and the heat lost by the iron is
{{{m*0.113*(40-x)}}} .
working in a Styrofoam cup the heat gained from or lost to the environment is negligible, so
{{{m*1*(x-20)=m*0.113*(40-x)}}} <--> {{{x-20=0.113*(40-x)}}} .
Solving:
{{{x-20=0.113*(40-x)}}}
{{{x-20=4.52-0.113x)}}}
{{{x+0.113x=20+4.52)}}}
{{{1.113x=24.52)}}}
{{{x=24.52/1.113)}}}
{{{x=22.03(rounded)}}}
 
B) It takes 79.72 calories to melt 1 gram of ice.
That is called enthalpy of fusion, or heat of fusion, or latent heat of fusion.
It is a lot of heat, so ice can take a lot of heat away from anything it touches.
Can 1 gram of ice take enough heat from 1 gram of water to cool it from 10 degrees to 0 degrees?
Let's see.
Cooling 1 gram of water from 10 degrees to 0 degrees (Celsius) takes only
{{{(1gram)*(10degrees)*1}}}{{{calorie/(gram*degree)=10degrees}}} .
So, if you mix 1 gram of water at {{{10^o}}}{{{C}}} and 1 gram of ice at {{{0^o}}}{{{C}}} ,
you get water at {{{0^o}}}{{{C}}} with a slightly smaller amount of ice floating in it, all of it at {{{0^o}}}{{{C}}} .