Question

A 218 kg rock sits in full sunlight on the edge of a cliff 5.05 m high. The temperature of the rock is 31.7 ∘C.

If the rock falls from the cliff into a pool containing 5.90 m3 of water at 16.5 ∘C, what is the final temperature of the rock-water system? Assume that the specific heat of the rock is 1010 J/(kg⋅K).

Answer 1

Given

rock of mass m1 = 218 kg , Temperature T1 = 31.7 0C

we know that 1 m^3 of water = 1000 kg ==> m2 = 5.90*1000= 5900 kg, T2 = 16.5 0C

specific heat of water is C2 = 4186 J/kg.k

specific heat of the rock is C1 = 1010 J/(kg⋅K).

here the heat lost by rock = heat gain by water  

we know that Q = m*C*dT

m1*C1(T-T1) = m2*C2(T- T2)

218*1010(T-31.7)= 5900*4186(T-16.5)

solving for T , T = 16.36 0C

the final temperature of the rock-water system is T =16.36 0C