A 275-g sample of nickel at 100.0°C is placed in 100.0 mL of water at 22.0°C. What is the final temperature of the water? Assume no heat is lost to the calorimeter. C(H₂O) = 4.18 J/(g.K), C(Ni) = 0.444 J/(g.K)

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A 275-g sample of nickel at 100.0°C is placed in 100.0 mL
of water at 22.0°C. What is the final temperature of the
water? Assume no heat is lost to the calorimeter.
C(H₂O) = 4.18 J/(g.K), C₂(Ni) = 0.444 J/(g.K)
Transcribed Image Text:A 275-g sample of nickel at 100.0°C is placed in 100.0 mL of water at 22.0°C. What is the final temperature of the water? Assume no heat is lost to the calorimeter. C(H₂O) = 4.18 J/(g.K), C₂(Ni) = 0.444 J/(g.K)
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Thank you. I'm just confused as what is is the reason 122.1 has to be multiplied by (-45543.23)?

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Heat lost by nickel = Heat gained by waterQNi = Qw m×CNi×(TNi – Tf) = mw×Cw×(Tf – Tw)275 g×0.444 Jg K×373-Tf=100 g×4.184 Jg K×Tf-295122.1×373-Tf=418.4×Tf-295373-Tf=3.4267×Tf-2953.4267 Tf-1010.8763 = 373-Tfsolve for Tf mathematicallyTf=312.62 K

This part is very confusing.

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