Question

A metal sample weighing 24.000 g is heated to 100.0 degrees celsius and then transferred into a calorimeter containing 30.0 mL of water at a temperature of 22.8 degrees celsius. If the specific heat of the metal is 0.105 J/g*C, what is the final temperature of the metal sample plus water?

Answer 1

Since density of water is 1 g/mL and volume is 30.0 mL,

m(water) = 30.0 g

T(water) = 22.8 oC

C(water) = 4.184 J/goC

m(metal) = 24.0 g

T(metal) = 100.0 oC

C(metal) = 0.105 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by metal = heat gained by water

m(metal)*C(metal)*(T(metal)-T) = m(water)*C(water)*(T-T(water))

24.0*0.105*(100.0-T) = 30.0*4.184*(T-22.8)

2.52*(100.0-T) = 125.52*(T-22.8)

252 - 2.52*T = 125.52*T - 2861.856

T= 24.3 oC

Answer: 24.3 oC