Question

100. mL of H2O is initially at room temperature (22.0∘C). A chilled steel rod at 2.0∘C is placed in the water. If the final temperature of the system is 21.1 ∘C, what is the mass of the steel bar? Specific heat of water = 4.18 J/g⋅∘C Specific heat of steel = 0.452 J/g⋅∘C Express your answer numerically, in grams, to one significant figure.

Answer 1

The quantity of heat content lost by water(at higher temperature) is equal to quantity of heat content gained by steel rod(lower temperature).

Q_{steel rod} = m Cv dT

m= mass of steel rod, = ?

Cv = specific heat of steel rod = 0.452 J/g⋅∘C

dT= differeence of temperatures 2.0^∘C and 21.1^∘C =  19.1^∘C

so Q_{steel rod} = m x 0.452 J/g⋅∘C x 19.1^∘C -----------------------------------------------------------------1

Qwater = m_w Cv dT

m_w= mass of water = 100 g ( 100 ml = 100 g) [density of water = 1g/ml]

Cv = specific heat of water = 4.18 J/g⋅∘C

dT= differeence of temperatures for water i.e final temperature - initial temperature,

21.1^∘C - 22.0^∘C

= -0.9^∘C

[initial tempeeature of water is 22.0^∘C and final temperature is 21.1^∘C, so

Qwater  = -(100 g x 4.18 J/g⋅∘C x -0.9^∘C)

= 376.2 J ----------------------------------------------------------------2

heat lost or given out by water , so  Qwater bears negative sign

equalize 1 and 2

m x 0.452 J/g⋅∘C x 19.1^∘C =  376.2 J   

mass of steel rod , m = 376.2 J / 0.452 J/g⋅∘C x 19.1^∘C

= 43.576 g

= 43.6 g (three significant figures) best mass report

= 43 g or 44 g (two signnificant figures) but not best report

to note as an inner point expressing in one significant figure not possible