Question

HW #3 Problem: An aluminum cup contains 225 g of water at 27°C. A 400g sample of silver at an
initial temperature of 56°C is placed in the water. A 40 g copper stirrer, that starts in
the water, is used to stir the mixture until the entire system reaches a final
temperature of 29°C. What is the mass of the aluminum cup?

Answer 1

Given mass of water m_w=225g,

Mass of silver m_Ag=400g, Mass of copper m_cu=40g

Mass of aluminum cup m_al=????

Also initial temperature of water and copper is T_w=T_cu=27°C

Initial temperature of silver is T_Ag=56°c

Now to find out the mass of aluminum cup we use conservation of heat energy.

That is Q of water+copper+aluminum=Qof silver

So as we know Q=mcΔT, where m=mass, c= specific heat, ΔT=change in temperature

Now total heat energy of aluminum+copper+water=Q

Therefore Q=(m_AlC_Al+m_wC_w+m_cuC_cu)ΔT

Where C of Al=0.900, C of copper=0.386, C of water=4.186

Therefore Q=[(m_Al×900+225×4186+40×386)(29-27)]/1000

Q=[m_Al×900+941850+15440](2)/1000

Now Q_Ag(of silver)=[400×233×(56-29)]/1000

Now according to conservation of energy these two energy should be equal.

Therefore Q=Q_Ag

[m_Al×900+941850+15440](2)/1000=[400×233×27]/1000

900m_Al=(2516400/2)-(941850+15440)

m_Al=334.34g