200 g of water at 30 degrees C is mixed with 20 g of silver at 350 degrees C, what's the final temp?
I'd really like to see all the steps so I can repeat this process with other questions on my homework.
This problem can be solved using the heat capacity.Inside the calorimeter, the temperature equilibrates. Meaning, when the water (30 degrees) is added to a hot silver (350 degrees) there will come a time that their temperature will be the same. That same temperature is called the final temperature.
Heat capacity has the formula
q = mc delta T
inside the calorimeter, we can assume that
qwater = -qsilver
applying the formula we can have;
(mc delta T)water = -(mc delta T)silver
where m= mass of the material
c = specific heat constant for the material
delta T is the change of temperature (in Kelvin)
We can substitute the values now. (remember we convert Celsius to Kelvin)
(200g)(4.186 J/gK)(Tf -303K) = - (20g)(.233J/gK)(Tf - 623K)
**for convenience we remove the units and we let Tf=x
200(4.186)(x-303) = - (20)(0.233)(x-623)
solve for x
837.2(x-303) = -4.66(x-623)
837.2x - 253671.6 = -4.66x + 2903.18
combine like terms
(837.2+4.66)x = 2903.18 + 253671.6
841.86x = 256574.78
x = 304.771 K or 31.8 degree Celsius