a solid of 150g is heated to 100c and dropped into a 500ml of water at 20c the equilibrium temperature of the mixture is how much?
Answers
Answer:
Firstly, let the final equilibrium temperature be T.
Now, the heat lost by the solid=the heat gained by the water.
(0.5)(150)(200-T)=(4.2)(400)(T-20)(0.5)(150)(200−T)=(4.2)(400)(T−20)
On either sides I manipulated the values to get Joules/gK and g. We are measuring temperature difference so we need not convert to Kelvin.
On operating we get,
T=\frac{648}{23.4}=27.7T=
23.4
648
=27.7
So the eqm temperature is 27.7 degrees Celsius.
The second part is a three step process, ice at -10 to ice 0 degrees Celsius,ice at 0 degrees celsius to water at 0 degrees Celsius and then water at 0 degrees Celsius to water at 80 degrees Celsius.
The energy absorbed by ice to get to 0 degrees is the energy needed to move up 10K or degrees Celsius,
E_{1}=(10)(40)(2.1)=840 JE
1
=(10)(40)(2.1)=840J
the energy needed for phase change is,
E_{2}=(320)(40)=12800JE
2
=(320)(40)=12800J
Here 320J/g is the latent heat of fusion of water.
E_{3}=(80)(4.2)(40)=13440JE
3
=(80)(4.2)(40)=13440J
Thus total energy needed is
q=E_{1}+E_{2}+E_{3}=27080Jq=E
1
+E
2
+E
3
=27080J
I'm too tired to do the rest xD