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**

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