Calculating Temperature Change of a Lead Ball Dropped from 106.0 m

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To calculate the temperature change of a lead ball dropped from 106.0 m, the energy from its fall is converted into heat, which raises its temperature. The relevant equation is Q = cm(ΔT), where Q is the energy lost, c is the specific heat capacity of lead, and m is the mass of the ball. The potential energy (PE) at the height can be calculated using PE = mgh, where g is the acceleration due to gravity. The mass can be canceled out in the calculations, allowing the temperature change to be determined solely from the height and specific heat capacity. Ultimately, the final temperature of the lead ball can be calculated using these principles.
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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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