Question

Try to solve this question

Don't spam ​

Answer 1

Heat energy imparted by calorimeter and water contained in it in cooling from 32°C to 5°C is used in melting ice and then raising the temperature of melted ice from 0°C to 5°C.

For cold body: Ice at 0°C

to water at 5°C. Heat gained = mL + mcAT

= m x 330 + m x 4.2 x (5-0)

= 330 m+m x 4.2 x 5

= 330m+21m = 351m joule.

For hot body:

(Water + Calorimeter) at 32°C to 5°C.

Heat lost = m₁c₁₁+ ₂₂ ₂

= 150 x 4-2 x (32-5)+ 50 x 0.4 x (32-5)

= 150 x 4.2 x 27 + 50 x 0-4 x 27

= 17010 + 540=17,550 joule.

From the principle of calorimetiy, if the system is fully insulated then, Heat gained by cold body = Heat lost by hot body.

$= > 351m = 17550 \\ = > m = \frac{17550}{351} \\ = > 50g$

Answer 2

Answer:

Given :-

Specific heat capacity of calorimeter = 0.4J/g°C

Specific heat capacity of water = 4.2J/g°C

Latent capacity of ice = 330J/g

Mass of Mw=150g

Specific heat capacity of water Sw= 4.2J/g°C

Initial temperature Ti= 32°C

Final temperature Tf = 5°C

Mass of calorimeter Mc= 50g

Specific heat capacity of calorimeter Sc=0.4J/g°C

Heat released Q= (Mw Sw + Ms Ss)

Q= (150×4.2+50×0.5) (32-5) = 17550J

$mass \: of \: ice \: requried \: mi = \frac{q}{l}$

Where L = 330J/g

$mi = \frac{17550}{330} = 53.2g$

mi= 53.2g

Please drop me some thanks please