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Incorrect Question 7 0/4 pts 63.0 g of an unknown metal at 100°C is placed in 50.0 g of water at 24.4°C. The temperature of t

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Answer #1

m(water) = 50.0 g
T(water) = 24.4 oC
C(water) = 1.0 cal/goC
m(metal) = 63.0 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))
63.0*C(metal)*(100.0-27.2) = 50.0*1.0*(27.2-24.4)
4586.4*C(metal) = 140
C(metal)= 0.0305 cal/goC
Answer: Pb(0.0305)

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