Question

2. A 10.0 g ice cube, initially at 0.0ºC, is melted in 100.0 g of water that was initially 20.0ºC. After the ice has melted, the equilibrium temperature is 10.93ºC. Show your work!
a. Calculate the total heat (in J) lost by the water (specific heat for water is 4.184 J/g°C).



b. Calculate the heat gained by the ice cube after it melts (specific heat for ice is 2.093 J/g°C).



c. Calculate the heat it took to melt the ice. (It takes 334.0 J/g of heat energy to melt ice). qmelt = mice ΔHmelt

Answer 1

Answer:

The heat lost  by the water  $Q_{water} =$ 3.8 KJ

The heat gain by ice  $Q_{ice}$ = 228.76 J

The heat required to melt the ice $Q_{melt}$ = 3340 J

Explanation:

Mass of ice cube $m_{ice}$ = 10 gm

Initial temperature of ice cube $T_{ice}$ = 0 °c

Mass of water $m_{water}$ = 100 gm

Initial temperature of water $T_{w}$ = 20 °c

Final temperature of mixture $T_{f}$ = 10.93 °c

(a). Total heat lost by the water $Q_{water} =$  $m_{water}$ $C_{w}$ ( $T_{f}$ - $T_{w}$ )

⇒ $Q_{water} =$ 100 × 4.184 (20 - 10.93)

⇒ $Q_{water} =$ 3.8 KJ

This is the heat lost  by the water.

(b). Heat gained by the ice cube $Q_{ice}$ = $m_{ice}$ $C_{ice}$ $(T_{f} - T_{ice} )$

⇒ $Q_{ice}$ = 10 × 2.093 × ( 10.93 - 0)

⇒ $Q_{ice}$ = 228.76 J

This is the heat gain by ice.

(C). Heat required to melt the ice $Q_{melt}$ = $m_{ice}$ × Latent Heat

⇒ $Q_{melt}$ = 10 × 334

⇒ $Q_{melt}$ = 3340 J

This is the heat required to melt the ice.