Question

A calorimeter containing first 50 g and then 80 g of water is heated and kept in same surrounding. It is found that the time to cool from 40°C to 30°C in the two cases was 28 and 43 minutes respectively. Water equivalent (in g) of the calorimeter is

Answer 1

Explanation:

50°C . . . 45° C . . . 4° C

Let the surrounding temperature be T∘C

Average T=50+452

952=47.5

Average temperature difference from surrounding

T=47.5−t

50−455=1∘C/m.

Rate of fall of temperature =5

From Newton's law

1∘C/mm=bA×T

⇒bA=1T=147.5−t

In 2nd case,

Average temperature

=40+452=852=42.5

Average temperature difference from surrounding

T'=42.5−t

Rate of fall of temperature

=45−408=(58)∘C/m

From Newton's law

58=bAT'

⇒58=42.5−t47.5−t

Hence, t=34.1∘C.

Hope it helps you !!