(a) Two 58 g ice cubes are dropped into 380 g of water in a thermally...

(a) Two 58 g ice cubes are dropped into 380 g of water in a thermally insulated container. If the water is initially at 22°C, and the ice comes directly from a freezer at -16°C, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used? The specific heat of water is 4186 J/kg·K. The specific heat of ice is 2220 J/kg·K. The latent heat of fusion is 333 kJ/kg.

Expert Solution

a)
Heat required to move ice from -16 oC to 0 oC
Q1 =m*Cice*delta T
= (2*0.058 Kg)* (2220 J/KgoC)*(16)
=4120.32 J

Heat released to move water from 22oC to 0 oC
Q2 =m*Cwater*delta T
= (0.38 Kg)* (4186 J/KgoC)*(22)
=34995 J

Heat required to melt all ice:
Q3 = m* Lice
= (2*0.058 Kg) * 333000
= 38628 J

Q1+Q3 = 42748.32 is less than Q2
So this heat can't be provided by water
So, final temperature = 0 oC
Answer: 0 oC

b)
Heat required to move ice from -16 oC to 0 oC
Q1 =m*Cice*delta T
= (0.058 Kg)* (2220 J/KgoC)*(16)
=2060.16 J

Heat released to move water from 22oC to 0 oC
Q2 =m*Cwater*delta T
= (0.38 Kg)* (4186 J/KgoC)*(22)
=34995 J

Heat required to melt all ice:
Q3 = m* Lice
= (0.058 Kg) * 333000
= 19314 J

Q1+Q3 = 21374.16 is less than Q2
this heat can be provided by water
Let final temperature be T oC
heat gained by ice = heat lost by water
2060.16 + 19314 + 0.058*4186*(T-0) = (0.38)* (4186)*(22-T)
2060.16 + 19314 + 242.788*T = 34994.96 - 1590.68*T
T = 7.43 oC

Answer: 7.43 oC

genius_generous answered 1 day ago

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