Calculate the energy that is required to change 50.0 g ice at -30.0°c to a liquid at 73.0°c. the heat of fusion = 333 j/g, the heat of vaporization = 2256 j/g, and the specific heat capacities of ice = 2.06 j/gk and liquid water = 4.184 j/gk.
1.31 × 105 j
2.14 × 104 j
1.66 × 104 j
3.50 × 104 j
6.59 × 103 j
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Ответ:
3,50x10⁴J
Explanation:
To obtain the energy required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C you need to caclulate:
Energy to increase temperature of ice from -30,0°C to 0,0°CHeat of fusion (Change of solid to liquid water)Energy yo increase the temperature of liquid water from 0,0°C to 73,0°C1. The energy is obtained with specific heat capacity of ice, thus:
Q = C×m×ΔT
Q = 2,06 J/gK×50,0g×(0,0°C- -30,0)
Q = 3090 J
2. The heat of fusion is:
333 J/g×50g = 16650J
3. Energy to increase temperature of liquid water is:
Q = C×m×ΔT
Q = 4,184 J/gK×50,0g×(73,0°C - 0,0°C)
Q = 15272 J
Thus, total heat is:
15272J + 16650J + 3090J = 35012J = 3,50x10⁴J
I hope it helps!
Ответ:
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Hope this helps! Good luck! :))