Answer to Question #97762 in General Chemistry for Brittany Wallace

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 chunk of tungsten weighing 19.40 grams and originally at 97.35 °C is dropped into an insulated cup containing 83.87 grams of water at 20.58 °C.
The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.83


J/°C.



Using the accepted value for the specific heat of tungsten (See the References tool), calculate the final temperature of the water. Assume that no heat is lost to the surroundings.



Tfinal = °C.

Answer 1

Q97596

Solution:

According to the Principle of Calorimetry ;

For a thermodynamic system , Heat given = Heat absorbed; provided no heat is lost to the surroundings.

Thus, $Q_{tungsten}=Q_{water}+Q_{calorimter}$ ---(1)

Also, at equilibrium final temperature of the system is the same for all it's constituents. Let this temperature be $T_{final }°C.$

$Q=ms\Delta T =c\Delta T$

where;

m = mass of the substance (in $gms.$ )

s = specific heat capacity (in $J/g°C$ )

$\Delta T=$ change in temperature of the substance (in $°C$ )

c= heat capacity of the substance (in $J/°C$ )

Given:

$m_{tungsten}=19.40 gms.$

$T_{i(tungsten)}=97.35°C$

$m_{water}=83.87gms.$

$T_{i(water)}=20.58°C = T_{i(calorimeter)}$ (since initially the calorimeter was in equilibrium with the water in it.

$c_{calorimeter}=1.83 J/°C$

Also;

$s_{water}=1J/g°C$

$s_{tungsten}=0.13J/g°C$

Thus; substituting values in (1), we get;

$19.40*0.13*(97.35-T_f)=83.87*1*(T_f-$$20.58)+1.83*(T_f-20.58)$

Solving for $T_f$ we get;

$2.522(97.35-T_f)=85.7(T_f-20.58)$

$2.522*97.35-2.522T_f=85.7T_f-85.7*20.58$

$245.517+1763.706=88.222T_f$

$T_f=2009.223/88.222$

$=22.775 ^\omicron C.$ (Answer)