Question

A silver block, initially at 56.4 ∘C, is submerged into 100.0 g of water at 24.6 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 27.4 ∘C.

What is the mass of the silver block?

Answer 1

Answer:

We know that heat, Q=mCpΔT=mCp(Tf-Ti)

where m=mass, Cp=specific heat, Tf=final temperature, and Ti=initial temperature.

Given the mass of water, m=100g, the initial temperature of the water, Ti=24.6∘C, Specific heat of water, Cp=4.186J/g∘C

and mass of silver block=??, Initial temperature of the silver block, Ti=56.4 ∘C, Specific heat of Silver block, Cp=0.233 J/g∘C,

Final temperature of the mixture, Tf=27.4 ∘C.

Now the heat lost by silver block=heat gained by water

-Qsilver block=Qwater

-(mCp(Tf-Ti))silver block=(mCp(Tf-Ti))water

-(m x 0.233 J/g∘C x (27.4∘C - 56.4 ∘C))= (100 g x 4.186 J/g∘C x (27.4∘C - 24.6∘C)

mass of silver block=(100g x 4.186 x 2.8)/(0.233 x 29)=173.46 g.

Therefore mass of silver block=173.46 g.

Please let me know if you have any doubt. Thank you