Question

Calculate the initial temperature of a 37.5 gram block of aluminum if the block of aluminum was submerged in 55.0 grams of water at 52.6°C and allowed to come to thermal equilibrium. The temperature of the system at thermal equilibrium was 40.5°C. (Assume the water and aluminum are thermally isolated from everything else.)

a) 12.1 °C

b) –41.7 °C

c) –33.9°C

d) –122 °C

e) 7.15°C

Answer 1

m(aluminium) = 37.5 g

T(aluminium) = to be calculated

m(water) = 55.0 g

T(water) = 52.6 oC

C(water) = 4.184 J/goC

T = 40.5 oC

We will be using heat conservation equation

use:

heat lost by water = heat gained by aluminium

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

55.0*4.184*(52.6-40.5) = 37.5*0.902*(40.5-T(aluminium))

T(aluminium)= -41.7 oC

Answer: b