Answer to Question #170436 in Chemistry for Jose Reyes

Question

A sample of nickel metal with a mass of 82.3 grams at 70.0 °C is dropped into 100.0 grams of water at 20.0 °C. If the specific heat of the nickel metal is 0.444 J/g°C and the specific heat of the water is 4.20 J/g°C, then what was the final temperature of the system?

Answer 1

Solution:

1) This problem can be summarized thusly:

q_{lost by nickel} = q_{gained by water}

q = m × C × ΔT

q = m × C × (T_f - T_i),

where:

q = amount of heat energy gained or lost by substance

m = mass of sample

C = specific heat capacity (J ^oC^-1 g^-1)

T_f = final temperature

T_i = initial temperature

2) Therefore:

q_{lost by nickel} = (82.3 g) × (0.444 J/g°C) × (T_f - 70.0)°C

q_{lost by nickel} = 36.5412 × (T_f - 70.0)

q_{gained by water} = (100.0 g) × (4.20 J/g°C) × (T_f - 20.0)°C

q_{gained by water} = 420 × (T_f - 20.0)

420 × (T_f - 20.0) = 36.5412 × (T_f - 70.0)

11.494 × (T_f - 20.0) = T_f - 70.0

11.494 × T_f - 229.878 = T_f - 70.0

10.494 × T_f = 159.878

T_f = 15.235°C = 15°C

Answer: The final temperature of the system (T_f) is about 15°C.