Question

A 275 g sample of nickel at 100.0ºC is placed in 100.0 mL of water at 22.0ºC. What is the final temperature of the mixture, assuming no heat is lost to the surroundings? The specific heat capacity of Ni is 0.444 J/(g.ºC). The density of water is 1.00 g/mL and the specific heat capacity of water is 4.18 J/(g.ºC.

Answer 1

m(H2O) = 100.0 g

T(H2O) = 22.0 oC

C(H2O) = 4.18 J/goC

m(Ni) = 275.0 g

T(Ni) = 100.0 oC

C(Ni) = 0.444 J/goC

Let the final temperature be T oC

use

heat lost by Ni = heat gained by H2O

m(Ni)*C(Ni)*(T(Ni)-T) = m(H2O)*C(H2O)*(T-T(H2O))

275.0*0.444*(100.0-T) = 100.0*4.18*(T-22.0)

122.1*(100.0-T) = 418*(T-22.0)

12210 - 122.1*T = 418*T - 9196

T= 39.6 oC

Answer: 39.6 oC

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