First slide

Calorimetry

Question

300 grams of water at $25^{\circ} C$ is added to 100 grams of ice at $0^{\circ} C$ , the final temperature of the mixture is

Moderate

A

$\frac{- 5}{3}^{\circ} C$
B

$- \frac{5}{2}^{\circ} C$
C

${-5}^{o} C$
D

$0^{\circ} C$

Solution

Water
${amount of heat lost by water is dQ}_{1} =mSdθ h e r e m i s m a s s o f w a t e r; S i s s p e c i f i c h e a t o f w a t e r; dθ i s c h a n g e i n t e m p e r a t u r e o f w a t e r w h e n w a t e r l o o s e s i t s t e m p e r a t u r e f r o m 25^{0} C t o 0^{0} C, a m o u n t o f h e a t l o s t i s {dQ}_{1}$
${dQ}_{1} =300 \times 1 \times 25=7500 calorie$
For Ice
${amount of heat required to convert from ice to water is dQ}_{2} =mL w h e r e m i s m a s s o f i c e; L = l a t e n t h e a t o f f u s i o n o f i c e = 80 c a l g^{- 1}$
${dQ}_{2} =100 \times 80 = 8000 cal$
${dQ}_{2} {>dQ}_{1} i.e., Heat required is greater than heat available$
Still ice remains= $0^{o} C$

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