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 at 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. Thermal temperature of the wax in ^ C is close to

Step-by-step solution

M _ B =20 10⁻³ Kg C _ B =5000 ~J / Kg - ^ C V =2000 M / s M _ w =1 Kg C _ w =3000 ~J / Kg - ^ C T _ f =25^ C =298 ~K 1 2 MV ²= M _ w C _ w T _ w + M _ B C _ B T _ B = 1 2 M _ B V ²= M _ w C _ w ( T _ w )+ M _ B C _ B T _ B 1 2 20 10⁻³ 4 10⁶ =( T ) 1 3000+20 10⁻³ 5000 40 10³= T 3000+100) T = 40 10³ 3100 T =12.9 ~T _ f -25=12.9 ~T _ f =25+12.9=37.9^ C