A 20 gm bullet whose specific heat is 5000 J-kg-C and moving at 2000 m-s plunges into a 1-0 kg block of wax whose specific heat is 3000 J-kg-C- Both bullet and wax are 25-C and assume that i the bullet comes to rest in the wax and ii all its kinetic energy goes into heating the wax- Find the Thermal temperature of the wax in -C-

Kinetic energy of the bullet, KE= 1 2 × 20 1000 × ( 200) ^ 2 J This KE goes to heating the wax. #160;Let after heating,temperature of the wax is t ...

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Standard XII

Chemistry

Kinetic Theory of Gases

A 20 gm bul...

Question

A $20 g m$ bullet whose specific heat is $5000 J / (k g^{\circ} C)$ and moving at $2000 m / s$ plunges into a $1.0 k g$ block of wax whose specific heat is $3000 J / (k g^{\circ} C)$ . Both bullet and wax are $25^{\circ} C$ and assume that (i) the bullet comes to rest in the wax and (ii) all its kinetic energy goes into heating the wax. Find the Thermal temperature of the wax in $^{\circ} C$ .

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Solution

Kinetic energy of the bullet,
$K E = \frac{1}{2} \times \frac{20}{1000} \times ({200)}^{2} J$
This KE goes to heating the wax.
Let after heating,temperature of the wax is $t ° C$ then,
$\frac{1}{2} \times \frac{20}{1000} \times ({200)}^{2} = 1 \times 3000 \times (t - 25) t = 38.33$
$∴$ temperature will be $38.33 ° C$

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Kinetic energy of the bullet,KE=12×201000×(200)2JThis KE goes to heating the wax. Let after heating,temperature of the wax is t°C then,12×201000×(200)2=1×3000×(t−25) t=38.33∴ temperature will be 38.33°C