1. (10 pts) A 150.0 g sample of a metal at 75.0 °C is added to 150.0 g H20 at 15.0 °C (s.h. = 4.184 J/g °C). The te...

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Question

Answer 1

Answer #1

m(water) = 150.0 g

T(water) = 15.0 oC

C(water) = 4.184 J/goC

m(metal) = 75.0 g

T(metal) = 150.0 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))

75.0*C(metal)*(150.0-18.3) = 150.0*4.184*(18.3-15.0)

9877.5*C(metal) = 2071.08

C(metal)= 0.2097 J/goC

Answer: 0.210 J/goC

Answer 2

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