A chunk of nickel weighing
18.0 grams and originally at
97.55 °C is dropped into an insulated cup
containing 81.0 grams of water at
23.10 °C.
Assuming that all of the heat is transferred to the water, the
final temperature of the water is ______ °C.
specific heat capacity of Nickel = 0.440 J/g.oC
m(water) = 81.0 g
T(water) = 23.1 oC
C(water) = 4.184 J/goC
m(nickel) = 18.0 g
T(nickel) = 97.55 oC
C(nickel) = 0.44 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))
18.0*0.44*(97.55-T) = 81.0*4.184*(T-23.1)
7.92*(97.55-T) = 338.904*(T-23.1)
772.596 - 7.92*T = 338.904*T - 7828.6824
T= 24.8001 oC
Answer: 24.80 oC
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