heat lost by an object of mass m and specific heat capacity s as temperature drops from t1 to t2 is given by
Answers
Answer:
A mass m
1
of a substance A of specific heat capacity c
1
at temperature T
1
is mixed with a mass m
2
of other substance B of specific heat capacity c
2
at a lower temperature T
2
and final temperature of the mixture becomes T.
Fall in temperature of substance A=T
1
–T
Rise in temperature of substance B=T–T
2
Heat energy lost by A=m
1
×c
1
× fall in temperature
= m
1
c
1
(T
1
–T)
Heat energy gained by B=m
2
×c
2
× rise in temperature
= m
2
c
2
(T–T
2
)
If no energy lost in the surrounding, then by the principle of mixtures,
Heat energy lost by A = Heat energy gained by B
m
1
c
1
(T
1
−T)=m
2
c
2
(T–T
2
)
After rearranging this equation, we get
T=
m
1
c
1
+m
2
c
2
m
1
c
1
T
1
+m
2
c
2
T
2
Here we have assumed that there is no loss of heat energy.