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Question 4 2.5 pts Following the procedure in lab, a student heated their 14.57 g metal sample to 100.0°C The metal was added
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Answer #1

m(water) = 54.849 g
T(water) = 29.5 oC
C(water) = 4.184 J/goC
m(metal) = 14.57 g
T(metal) = 100.0 oC
C(metal) = to be calculated

We will be using heat conservation equation

use:
heat lost by metal = heat gained by water
m(metal)*C(metal)*(T(metal)-T) = m(water)*C(water)*(T-T(water))
14.57*C(metal)*(100.0-37.5) = 54.849*4.184*(37.5-29.5)
910.625*C(metal) = 1835.9057
C(metal)= 2.0161 J/goC
Answer: 2.02 J/goC

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