Question

A 306.4-milligram cube of iron is heated to 1,423°C and placed in a calorimeter filled with77.1-mL of 25.0°C air. Calculate the temperature at thermal equilibrium. The density of air is 1.225 x 10-3 g/cm3.

Answer 1

Specific heat of iron (C_{​​​​​​iron} ) = 0.444 J/g°C

Specific heat of air (C_{​​​​​​air}) = 1.020 J/g°C

T_{​​​​​​Initial} of Iron = 1423°C

Mass of iron = 306.4 mg = 0.3064 g

T_{​​​​​​Initial} of air = 25°C

1 mL = 1 cm³

Mass of air = Density × Volume = (1.225 × 10^-3 g/mL) × 77.1 mL = 94.44 g

So, Heat lost by iron = -Heat absorbed by air

​​​​​​=> mC_{​​​​​Iron}​∆T = -mC_{​​​​​air}​​∆T

=> (0.3064 g)*(0.444 J/g°C)*(1423 - T)°C = (94.44 g)*(1.020 J/g°C)*(T - 25)°C

=> 193.6 - 0.136 T = 96.33 T - 2408.4

=> T = 2602/96.466 ≈ 27°C

Hence, 27°C is correct answer.