an iron ball weighing 100 gm and heated to 98.5degree C are dropped in a calorimeter weighing 46gm and containing 85.4 gm of water at 15degree C. the final temperature of the mixture becomes 22degree C. Calculate the specific heat of iron (Given specific heat of calorimeter material = 0.1cal/gm)
Answers
Answer:
Given: A piece of iron weighing 100g was heated to 98.5
o
C and dropped in a calorimeter weighing 46g and containing 85.4g water at 15
o
C. The temperature was found to be 22
o
C.
To find the specific heat of iron if the specific heat of material of calorimeter is 0.1.
Solution:
As per the given criteria,
Mass of iron, m
i
=100g
Mass of the water in calorimeter, m
w
=85.4g
Water equivalent of the calorimeter, w
e
=46×0.1=4.6g
Temperature of iron, T
i
=98.5
∘
C
Temperature of water or calorimeter, T
w
=15
∘
C
Final temperature, T=22
∘
C
Heat lost by the iron = mass\times specific heat\times fall in temperature
H
L
=msΔT
⟹H
L
=100×s×(98.5−22)=7650s
Heat gained by the calorimeter,
H
g
=msΔT⟹H
g
=4.6×1×(22−15)
⟹H
g
=32.2cal
Heat gained by water,
H
w
=msΔT
⟹H
w
=85.4×1×(22−15)
⟹H−w=597.8cal
Therefore total heat gained = 32.2+597.8=630cal
According to the principle of calorimetery,
Heat lost=Heat gained
7650s=630
⟹s=
7650
630
⟹s=0.0823
is the required specific heat of iron