a) A 500.-g iceberg at -5.00°C is placed in a 226 g beaker with specific heat...

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?

Solutions

Expert Solution

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

genius_generous answered 1 year ago

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