Answer to Question #136699 in General Chemistry for qwerty

Question

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction.

A student heats 61.00 grams of lead to 99.19 °C and then drops it into a cup containing 79.31 grams of water at 24.32 °C. She measures the final temperature to be 26.53 °C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.53 J/°C.

Assuming that no heat is lost to the surroundings calculate the specific heat of lead.

Answer 1

Mass of lead, M = 61.00 grams

= 61× 10^-3 kg

=0.061 kg

Initial temperature, T₁ = 99.19 °C

After dropping it in calorimeter,

the final temperature of whole system, T₂ = 26.53 °C.

Let,

heat capacity of lead = S J.kg^-1.°C^-1

Heat liberated by lead, H = M×S×(T₂-T₁)

Or, H = 0.061× S J.kg^-1.°C^-1 × ( 99.19 - 26.53 )°C

Or, H = 0.061×S×72.66 Joul

Or, H = 4.432×S Joul

Mass of water in the calorimeter, M' = 79.31 grams

M' = 79.31×10^-3 kg

= 0.07931 kg

Initial temperature of water in calorimeter, T₁' = 24.32 °C.

The final temperature of whole system, T₂' = 26.53 °C.

Heat capacity of water = 1 Cal g^-1 °C^-1

1Cal = 4.2 Joul and 1 g = 10^-3 kg

So,

Heat capacity of water, S' = 4.2J × (10^-3 kg)^-1 × °C^-1

= 4200 J kg^-1 °C^-1

Heat gain by water, H' = M'×S'×(T₂' - T₁')

Or, H' = 0.07931 kg×4200 J kg^-1 °C^-1×(26.53 - 24.32) °C

Or, H' = 736.155 J

As the calorimeter is in thermal equilibrium with the water in it before and after.

So,

Initial temperature of the calorimeter = T₁

= 24.32 °C

The final temperature of whole system, T₂= 26.53 °C.

Heat capacity of calorimeter, S" = 1.53 J °C^-1

Heat gain by the calorimeter,

H" = S"×(T₂ - T₁)

Or, H" = 1.53 J °C^-1 × (26.53 - 24.32) °C

Or, H" = 3.381 Joul

Heat liberated by lead, H = (Heat gain by water, H' + Heat gain by the calorimeter,H")

I.e. H = H' + H"

Or, 4.432×S = 736.155 + 3.381

Or, S = (736.155 + 3.381) / 4.432

= 166.862

Hence, heat capacity of lead is 166.862 Joul kg^-1 °C^-1