Specific Heat Capacity A 21.5-g sample of an unknown metal is heated to 94.0°C and is...

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Question

Answer 1

Answer #1

m(water) = 128.0 g

T(water) = 21.4 oC

C(water) = 4.18 J/gK = 4.18 J/goC

m(metal) = 21.5 g

T(metal) = 94.0 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))

21.5*C(metal)*(94.0-25.0) = 128.0*4.18*(25.0-21.4)

1483.5*C(metal) = 1926.144

C(metal)= 1.2984 J/goC

C(metal)= 1.2984 J/gK

Answer: 1.30 J/gK

Answer 2

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