Calculating Temperature Change of a Lead Ball Dropped from 106.0 m

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The discussion focuses on calculating the temperature change of a lead ball dropped from a height of 106.0 m, with an initial temperature of 25 °C. The relevant formula is Q = mcΔT, where c is the specific heat capacity of lead at 128 J/(kg·°C). The kinetic energy of the ball upon impact converts entirely into thermal energy, allowing for the calculation of the final temperature after the fall. The mass of the ball cancels out in the calculations, simplifying the process.

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A lead ball, with an initial temperature of 25 °C, is released from a height of 106.0 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 in °C of the ball after it hits. (You do not need to enter the units.) Data: c of lead = 128 [(J)/(kg·° C)].

I don't know where to go in this problem

I know you use Q=cm(deltaT) But I don't know what the mass and Q are so where do I go with this equation? Please help me out!
 
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Maybe the mass cancels out... What can you say about the kinetic energy of the ball as a function of the mass...
 

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