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vodka [1.7K]
3 years ago
11

A student melts 50.0 grams of ice using 16650 J of energy. What is the heat of fusion of ice?

Chemistry
2 answers:
san4es73 [151]3 years ago
8 0

Answer:

L = 330.0 J/g

Explanation:

The heat of fusion is the energy necessary to transform 1 gram of the solid compound in its liquid phase. Fusion requires that the systems gain heat, so it's an endothermic process.

The energy (Q) is relacionated to the mass of the compound (m) and its heat of fusion (L):

Q = mL

16650 = 50.0L

L = 16650/50

L = 333.0 J/g

Shkiper50 [21]3 years ago
6 0
Heat fusion is the amount of energy required to convert one mole of ice into water, 16650J is per 50g so if we divide 16650 by 50 we find it per gram: 16650/50 = 333J/g or 0.33KJ/g

hope that helps :)
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Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

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The balanced chemical reaction is,

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

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