80 g of water at $60^{o} C$ is poured into a vessel containing 100 g of water at $20^{o} C$ .The final temperature recorded is $30^{o} C$ . Calculate the thermal capacity of the vessel.

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Solution

For Hot water:
$M a s s = 80 g$
$S . H . C = 4.2 J g^{- 1}^{o} C^{- 1}$
$I n i t i a l T e m p . = 60^{o} C$
$F i n a l T e m p . = 30^{o} C = θ_{F} = (60 - 30) = 30^{o} C$
for Cold water:
$M a s s = 100 g$
$S . H . C = 4.2 J g^{- 1}^{o} C^{- 1}$
$I n i t i a l T e m p . = 20^{o} C$
$F i n a l T e m p . = 30^{o} C$
$= θ_{R} = (30 - 20)$
$= 10^{o} C$
for Vessel:
$M a s s = ? (m)$
$S . H . C = ? (c)$
$I n i t i a l T e m p . = 20^{o} C$
$L e t t h e r m a l c a p a c i t y o f v e s s e l = x = m c$
$H e a t l o s t b y h o t w a t e r = m c θ_{F}$
$= 80 \times 4.2 \times 30$
$= 10080 J$
$H e a t g a i n e d b y c o l d w a t e r = m c θ_{R}$
$= 100 \times 4.2 \times 10$
$= 4200 J$
$H e a t g a i n e d b y v e s s e l = m c θ_{R}$
$= x \times 10^{o} C$
$H e a t g a i n e d = H e a t l o s t$
$4200 + (x \times 10) = 10080 J$
$x = \frac{(10080 - 4200)}{10} J^{o} C^{- 1}$
$= 588 J^{o} C^{- 1}$

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