A 206 kg rock sits in full sunlight on the edge of a cliff 5.70 m high. The temp

Question

A 206 kg rock sits in full sunlight on the edge of a cliff 5.70 m high. The temperature of the rock is 30.20°C. If the rock falls from the cliff into a pool containing 6.00 m3 of water at 6.70°C, what is the final temperature of the rock-water system? Assume that the specific heat of the rock is 1010 J/(kg·K). Express your answer with three significant digits.

Explanation / Answer

This is a two-part question. 1. Find the temp. due to the mixture. 2. Add the temp. change due to the KE of the falling rock. 1. c1 = 1010 J/kg-K, c2 = 4186 J/kg-K m1 = 206 kg, m2 = 6000 kg T1 = 30.2 C, T2 = 15.5 C Basic equation is c1m1(T3-T1) = -c2m2(T3-T2) ==> T3 = (m1c1T1+m2c2T2)/(m1c1+m2c2) = 15.6208 C m3 = m1+m2 = 6206 kg c3 = (m1c1+m2c2)/m3 = 4080.577 J/kg-K 2. KE = mgh = 415872.8 J Tfinal = T3 + KE/(m3*c3) = T3 + 0.0164 = 15.6372 C

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