In: Physics
a) A 500.-g iceberg at -5.00°C is placed in a 226 g beaker with specific heat capacity 840 J/kg °C at temperature 15.0°C that contains 220 g of water. What is the minimum amount of steam needed to add into the system in order to melt the iceberg?
Energy required by the ice to completely melt, Qi = mCdT +
mLf
Where m is the mass of ice, C is the specific heat of ice and Lf is
latent heat of fusion of ice.
Qi = 500 x 2.04 x (5 - 0) + 500 x 334
= 172100 J
Energy given out as heat when glass and water is cooled to zero
degree
Qf = Mw Cw dT + Mg Cg dT
= 220 x 4.186 x 15 + 226 x 0.84 x 15
= 16661.4 J
Cooling of glass and water takes off 16661.4 J of energy from
ice
Remaining energy = 172100 J - 16661.4 J
= 155438.6 J
155438.6 = Mv Lv + Mv Cw dT
Where Mv and Lv are the mass and specific heat of vapour
= Mv [2230 + 4.186 x 100]
= Mv [2648.6]
Mv = 155438.6 / 2648.6
= 58.69 g