First slide

Heat ;work and internal energy

Question

A stationary object at 4° C and weighting 3.5 kg falls from a height 2000 metres on snow mountain at 0° C. If the temperature of the object at the strike is 0° C (g=10 $m s^{- 2}$ ), then amount of ice melted is:(Latent heat of fusion of ice=3.5× $10^{5}$ J/kg)

Moderate

A

20 grams
B

100 grams
C

200 grams
D

2000 grams

Solution

$\begin{array}{rcl} F r o m J o u l e s l a w : W & = & J Q \Rightarrow m_{o b j e c t} g h = m_{i c e m e l t e d} L \end{array}$
$\begin{array}{rcl} (I n i t i a l l y b o d y h a s p o t e n t a l e n e r g y i . e w o r k i s s t o r e d i n t h e f o r m o f p o t e n t e n t i a l e n e r g y & d u r i n g c o n v e r s i o n o f s t a t e, Q = m L) \end{array}$
$\begin{array}{rcl} H e r e m i s m a s s \\ g i s a c c e l e r a t i o n d u e t o g r a v i t y & = & 10 m s^{- 2} \\ h i s h e i g h t o f f a l l & = & 2000 m \\ L i s l a t e n t h e a t o f f u s i o o f i c e & = & 3.5 x 10^{5} J k g^{- 1} \\ s u b s t i t u t e a b o v e v a l u e s, (3.5) (10) (2000) & = & m_{i c e} (3.5 x 10^{5}) \\ m_{i c e} & = & \frac{(3.5) (10) (2000)}{3.5 x 10^{5}} k g \\ \Rightarrow & m_{i c e} = \frac{(3.5) (10) (2000)}{3.5 x 10^{5}} x 1000 g \\ \Rightarrow & m_{i c e} = 200 g = m a s s o f i c e t h a t m e l t e d \end{array}$

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