What is the final temperature of the silver?

Cs(Ag)=0.2350J/g*C Cs(H2O)=4.180J/g*C

Assuming the final temperature of both silver and water is T.

Then the heat silver lost=50.00(75.00-T)0.2350 J

The heat water absorbed=100.0(T-22.00)4.180 J

Therefore 50.00(75.00-T)0.2350=100.0(T-22.00)4.180

11.75(75.00-T)=418.0(T-22.00)

881.25-11.75T=418.0T-9196

429.75T=10077.25

T=23.45 C

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