A block of silver (SH=0.057cal/gC) weighing 55.0g was heated to a temperature of 98.4C. It was...
A block of silver (SH=0.057cal/gC) weighing 55.0g was heated to a temperature of 98.4C. It was transferred to a calorimeter containing 100.0g of water. The maximum temperature experienced by the calorimeter system was 26.5C. You realize you didn’t take the initial temperature of the water. Assuming complete heat transfer, can you calculate the initial temperature of the water?_________
80. If so, what is the value?_______________
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Homework Answers
m(water) = 100.0 g
T(water) = to be calculated
C(water) = 1 cal/g.oC
m(silver) = 55.0 g
T(silver) = 98.4 oC
C(silver) = 0.057 cal/goC
T = 26.5 oC
We will be using heat conservation equation
use:
heat lost by silver = heat gained by water
m(silver)*C(silver)*(T(silver)-T) = m(water)*C(water)*(T-T(water))
55.0*0.057*(98.4-26.5) = 100.0*1.0*(26.5-T(water))
T(water)= 24.2 oC
Answer: Yes, 24.2 oC
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