# The weight falls to the ground and hits an obstacle. The speed of the weight before impact is 140 m / s.

**The weight falls to the ground and hits an obstacle. The speed of the weight before impact is 140 m / s. What was the temperature of the weight before the impact, if after the impact the temperature increased to 100 ° C. Consider that the entire amount of heat released during the impact is absorbed by the weight. The specific heat capacity of the weight is 140 J / kg**

To find out the initial temperature of the specified weight, consider the equality: 0.5 * m * (V2 ^ 2 – V1 ^ 2) = A (work) = Q (heat) = Cr * m * (t2 – t1), whence t1 = t2 – 0.5 * m * (V2 ^ 2 – V1 ^ 2) / (Cr * m) = t2 – (V2 ^ 2 – V1 ^ 2) / 2Cr.

Data: t2 – final temperature of the weight (t2 = 100 ºС); V2 – speed before impact (V2 = 140 m / s); V1 – initial speed of the weight (V1 = 0 m / s); Cr is the specific heat capacity of the weight material (Cr = 140 J / (kg * K)).

Let’s perform the calculation: t1 = t2 – (V2 ^ 2 – V1 ^ 2) / 2Cg = 100 – (1402 – 02) / (2 * 140) = 30 ºС.

Answer: The initial temperature of the indicated weight was 30 ºС.