A 13.50 g sample of Au is heated, then placed in a beaker containing 60.00 g water. The temperature of water increases from 19.00 ºC to 20.00 ºC. What is the initial temperature of the gold sample? Cs (Au) = 0.1300 J/gºC; Cs (H2O) = 4.184 J/gºC
m(water) = 60.0 g
T(water) = 19.0 oC
C(water) = 4.184 J/goC
m(gold) = 13.5 g
T(gold) = to be calculated
T = 20.0 oC
We will be using heat conservation equation
use:
heat lost by gold = heat gained by water
m(gold)*C(gold)*(T(gold)-T) = m(water)*C(water)*(T-T(water))
13.5*0.13*(T(gold)-20.0) = 60.0*4.184*(20.0-19.0)
1.755*(T(gold)-20.0) = 251.04
T(gold)= 163 oC
Answer: 163 oC
A 13.50 g sample of Au is heated, then placed in a beaker containing 60.00 g...
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