Question

Assume you are performing the calibration step of Experiment 8 and you begin with 30 g of water at 20 °C and 30g of water at 80 °C. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 °C. What is the heat capacity of the calorimeter?

Answer 1

The calorimeter absorbes this heat cooling the mixture from 50 to 45°c.so the heat capacity is 1260/5=25/J/k

q(warm water)=q(cool water)+q(calorimeter) specific heat(warm water)×mass(warm water)×delta temp

Specific heat(cool water)×mass(cool water)×delta temperature

First start by substraction q(cool water)over the equal sign which will

-q(warm water)-q(cool water)=q(calorimeter)

Specific heat of water=4.184(J/grams)c(celsius)

C(calorimeter)×(45-25)-4.184J/gram)×70gms×45-20-(4.184(J/grams)×(70gms)×(45-80)

The first change of ten.(45-25)in for the calorimeter.H basically means that the calorimeter starts at25°cand ends at 45°c

After the warm and cold mixture

C( calorimeter)×(20c(celsius)=2,928.8

C,(calorimeter)=146.44J/C(celsius)

Now to covert to call/C( celsius)

146.44J/C(celsius)×(1cal/4.18400J)=35cal/C(celsius)

Heat capacity is calorimeter is usually in joule per Kelvin when you mix the two water sample together the temperature should since,the masses are the same be halfway between the two,or 50°c

The energy stored is referenced to 45°c is specific heat of water is 4.186KJ/kg

E= 4.186KJ/kgC×0.06kg×5°=1.26KJ=1260J