A volume 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