A lead ball, with an initial temperature of 25°C is released from a height of 12

Question

A lead ball, with an initial temperature of 25°C is released from a height of 121.0 m. It does not bounce when it hits a hard surface. Assume all the energy of the fall goes into heating the lead. Find the temperature of the ball after it hits. Data: clead = 128 J/(kg·°C). (Give your answer in Kelvin i.e., K.)

Explanation / Answer

potential energy,p.e =mgh where m = mass of lead,h= height of fall g=acceleration due to free fall. p.e=m(81)(9.8)=> p.e=793.8mJ this heat must go to cause temperature change in lead since and imperfect collision the ball has with the ground and it is assume that it in invested in the ball as heat energy. (potential energy converted to heat)=(heat that cause temperature rise in the lead ball,Q) BUT Q=(mass x specific heat capacity of lead x temperaturechange,T) Q=mcT Q=m(128)(@-25) but since Q=p.e=793.8mJ =>793.8m=m(128)(t-25) 793.8=128(t-25) (t-25)=(793.8/128) t-25=6.2 =>t=6.2+25 t=31.2°C = 304.2K therefore after heating the ground the ball has a temperature rise of 6.2°C as a result of p.e converted to heat energy giving it a final temperature of 31.2°C, i.e., 304.2 K

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