Question

110 mL of H2O is initially at room temperature (22∘C). A chilled steel rod at 2∘C is placed in the water. If the final temperature of the system is 21.3 ∘C, what is the mass of the steel bar?

Answer 1

we know that

for water

density = 1 g / ml

now

mass = density x volume

so

mass of water = 1 x 110

mass of water = 110 g

information required :

specific heat of water = 4.184 J / g C

specific heat of steel = 0.45 J / g C

now

we know that

heat lost by hot body = heat gained by cold body

in this case

hot body is water

and cold body is steel

so

heat lost by water = heat gained by steel

also

heat = mass x specific heat x temp change

so

m x s x dT for water = m x s x dT for steel

so

110 x 4.184 x ( 22 - 21.3) = m x 0.45 x ( 21.3 - 2)

solving we get

m = 37.1

so

the mass of steel bar is 37.1 grams