Avolume of 85.0 ml of h2o is initially at room temperature (22.00 âc). a chilled steel rod at 2.00 âc is placed in the water. if the final temperature of the system is 21.50 âc , what is the mass of the steel bar?

A heat vanished by the water is equivalent to heat gained by the cold rod 
Heat lost Q = mcΔT
Where:
Q is the quantity of heat,
m = mass of water in g,
c is specific heat in J, g^-1 ^-deg (= 4.18 for H2O) and
ΔT is the temperature change. 
Q = 85.0 × 4.18 × (22.0 - 21.5) = 177.65 J 
Heat gained steel rod Q = mcΔT = m × 0.452 × (21.5 -2.00) = 8.814 m 177.65 = 8.814 m 
m = 20.16 g 

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