Question

If hot water loses 750 cal while cold water absorbs 544 cal when the temperature change is 37.2 C for the cold water and the calorimeter, what is Ccal, the calorimeter constant?

22.154 g of cold water at 19C rises to a temperature of 32 C in a calorimeter with a Ccal of 5 cal/g C. 48.97 grams of Cu is initially at 100 C and reaches 32 C, the same as the cold water and the calorimeter. Determine the Csp of Cu according to this data?

Answer 1

Calculation of Ccal (calorimeter constant)

Calorimeter constant is the amount of heat required to increase the temperature of calorimeter by 1°C

Here the heat lost by hot water is partly absorbed by the cold water and partly by the calorimeter

Heat lost by hot water=heat gained by cold water+heat gained by calorimeter

750 cal=544 cal+ Ccal x rise in temperature

750 cal=544 cal+ Ccal x 37.2°C

750 cal-544 cal=Ccal x 37.2°C

206 cal=Ccal x 37.2°C

206 cal/37.2°C = Ccal

Ccal=5.54 cal/°C

Calculation of Csp of Cu (specific heat capacity of copper metal)

In this question the copper metal at 100°C loses heat which gets absorbed by cold water at 19°C and the calorimeter

So heat lost by copper =heat gained by cold water + heat gained by calorimeter

Mass of copper x specific heat of copper x fall in temperature

=Mass of cold water x specific heat of water x rise in temperature+ calorimeter constant x rise in temperature

48.97 g x Csp x (100°C -32°C)= 22.154 g x 1 cal/g°C x (32°C-19°C) + 5 cal/g

48.97 g x Csp x 68°C= 22.154 g x 1 cal/g°C x 13°C + 5 cal/g x 13°C

=288.002 cal + 65 cal

=353.002 cal

Csp=353.002 cal/(48.97 g x 68°C)=0.106 cal/g°C

So specific heat of copper according to this data=Csp=0.106 cal/g°C