First slide

Heat ;work and internal energy

Question

A lead ball moving with a velocity V strikes a wall and stops. If 50% of its energy is converted into heat, then what will be increase in temperature? (specific heat of lead is S)

Moderate

A

$\frac{{2V}^{2}}{JS}$
B

$\frac{V^{2}}{4JS}$
C

$\frac{V^{2} S}{J}$
D

$\frac{V^{2} S}{2J}$

Solution

$\begin{array}{rcl} f r o m J o u l e s l a w \\ w o r k d o n e i s c o n v e r t e d t o h e a t e n e r g y & \Rightarrow & W = J Q - - - (1) \\ 50 % o f m o v i n g e n e r g y & = & \frac{50}{100} x K E = W (f r o m W = ∆ K E) \\ W & = & \frac{50}{100} x \frac{1}{2} m V^{2} - - - (2) \\ h e r e m i s m a s s o f t h e b a l l; V i s v e l o c i t y o f b a l l \\ h e a t e n e r g y Q & = & m S d θ - - - - (3) \\ h e r e S i s s p e c i f i c h e a t; d θ i s c h a n g e i n t e m p e r a t u r e \\ s u b s t i t u t e e q u a t i o n s (2), (3) i n (1) \\ \frac{1}{2} (\frac{1}{2} m V^{2}) =JmSdθ \\ \Rightarrow & dθ= \frac{V^{2}}{4JS} \\ increase in temperature & = & \frac{V^{2}}{4JS} \end{array}$

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