Question

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 that no heat is lost to or gained from the surroundings. Specific heat capacity of nickel = 0.444 J/(g·K). a. 40.8°C b. 61.0°C c. 39.6°C d. 82.4°C e. 79.2°C

Answer 1

Since density of water is 1 g/mL and volume is 100.0 mL,

m(water) = 100.0 g

T(water) = 22.0 oC

C(water) = 4.184 J/goC

m(Nickel) = 275.0 g

T(Nickel) = 100.0 oC

C(Nickel) = 0.444 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by Nickel = heat gained by water

m(Nickel)*C(Nickel)*(T(Nickel)-T) = m(water)*C(water)*(T-T(water))

275.0*0.444*(100.0-T) = 100.0*4.184*(T-22.0)

122.1*(100.0-T) = 418.4*(T-22.0)

12210 - 122.1*T = 418.4*T - 9204.8

T= 39.6204 oC

Answer: 39.6 oC