Calculate the amount of ice, which is sufficient to cool 150 g of water, contained in a vessel of mass 50 g (S.H.C. $0.4 J g^{- 1}^{o} C^{- 1}$ ) at $30^{o} C$ , such that the final temperature of the mixture is $5^{o} C$

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Solution

For Ice:
$M a s s = ? (x)$
$S . H . C . / S . L . H . = 336 J g^{- 1}$
$I n i t i a l T e m p . = 0^{o} C$
$F i n a l T e m p . = θ_{R} = (5 - 0) = 5^{o} C$
For Water:
$M a s s = 150 g$
$S . H . C . / S . L . H . = 4.2 J g^{- 1}^{o} C^{- 1}$
$I n i t i a l T e m p . = 30^{o} C$
$F i n a l T e m p . = θ_{F} = (30 - 5) = 25^{o} C$
For Vessel:
$M a s s = 50 g$
$S . H . C . / S . L . H . = 0.4 J g^{- 1}^{o} C^{- 1}$
$I n i t i a l T e m p . = 30^{o} C$
$H e a t g a i n e d b y i c e t o f o r m w a t e r a t 0^{o} C = m L_{i c e} = x \times 336$
$H e a t g a i n e d b y w a t e r f o r m e d f r o m i c e = m c θ_{R} = x \times 4.2 \times 5 = 21 x$
$∴ T o t a l h e a t g a i n e d = 336 x + 21 x$
$= 357 x$
$H e a t l o s t b y w a t e r a t 30^{o} C = m c θ_{F}$
$= 150 \times 4.2 \times 25$
$= 15750 J$
$H e a t l o s t b y v e s s e l a t 30^{o} C = m c θ_{F}$
$= 30 \times 0.4 \times 25 = 300 J$
$T o t a l h e a t l o s t = (15750 + 300)$
$= 16050 J$
$H e a t g a i n e d = H e a t l o s t$
$357 x = 16050$
$∴ x (M a s s o f i c e) = \frac{16050}{357}$
$= 44.95 g$

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