ICSE Class 10 Physics 2022 Solved Paper

Question : 22

Total: 45

A metal piece of mass 420g present at 80∘C is dropped in 80g of water present at 20∘C in a calorimeter of mass 84g. If the final temperature of the mixture is 30∘C then calculate the specific heat capacity of the metal piece. [Specific heat capacity of water =4.2Jg−1‌∘C−1.
(Specific heat capacity of the calorimeter = 200Jkg−1‌∘C−1)

Solution:

Given:
Mass of metal piece (m1)=420g
Spécific heat capacity of metal piece (C1)= ?
Initial temperature of metal piece (T1)=80∘C
Mass of water (m2)=80g
Specific heat capacity of water (C2)
=4.2Jg−1‌∘C−1
Initial temperature of water (T2)=20∘C
Mass of calorimeter (m3)=84g
Specific heat capacity of calorimeter
‌=200Jkg−1‌∘C−1
‌=‌

200

1000

Jg−1‌∘C−1
‌=0.2Jg−1‌∘C−1
Final temperature of mixture =30∘C
According to the principal of calorimetry,
Heat lost by metal piece (hot body) = Heat gained by water
+ Heat gained by calorimeter (cold body)
‌m1C1∆T1=m2C2∆T2 +m3C3∆T3
‌420×C1×(80−30)=80×4.2×(30−20)
+84×0.2×(30−20)
‌21000×C1=3360+168
‌21000C1=3528
‌C1=‌

3528

21000

‌‌=0.168Jg−1‌∘C−1

Go to Question:

Prev Question

Next Question