A 275-g sample of nickel at 100.0 degree C is placed in 100.0 mL of water at 22.

Question

A 275-g sample of nickel at 100.0 degree C is placed in 100.0 mL of water at 22.0 degree C, the final temperature of the water? Assume that no heat is lost to or gained from the surroundings. Specific heat capacity of nickel = 0.444 J/(g middot K) A) 39.6 degree C B) 40.8 degree C C) 61.0 degree C D) 79.2 degree E) 82.4 degree C What is the energy in joules of a mole of photons associated with red light of 7.00 times 10^2 nm (c = 3.0 times 10^8 m/s; h = 6.63 times 10^-34 J-s. n_A = 6.022 times 10^23/mole? A) 256 kJ B) 1.71 times 10^5 J C) 4.72 times 10^43 J

Explanation / Answer

answer : A) 39.6 oC

mass of nickel = 275 g

temperature = 100 oC

mass of water = 100 mL = 100g

temperature of water = 22.0 oC

here :

heat loss by nickel = heat gain by water

(m Cp dT ) nickel = (m Cp dT)water

275 x 0.444 x (100 - Tf) = 100 x 4.184 x (Tf - 22)

122.1 (100 - Tf) = 418.4 (Tf - 22)

100 - Tf = 3.427 Tf - 75.39

Tf = 39.6oC

final temperature = 39.6 oC

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