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3 years ago

What is the temperature at which something boils? A chemical change or property or a physical change or property?

1 answer:

8 0

The boiling point of a substance is a physical property. A physical property of a material or substance is one that can be observed without changing or altering the composition of the material. Examples are mass, Density, Color, solubility, boiling point, melting point . A chemical property of a substance is one that describes how the material changes into a completely different substance and is observed only during a chemical reaction. Examples of chemical properties include types of chemical bonds, heat of combustion, reactivity with other metals, oxidation state and enthalpy of formation.

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2 years ago

Ethanol (C2H5OH) melts a - 144 oC and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kj/mol, and its enthalpy of vapo

hammer [34]

Answer:

For a: The total heat required is 36621.5 J

For b: The total heat required is 58944.5 J

Explanation:

For a:

To calculate the heat required at different temperature, we use the equation:

$q=mc\Delta T$ .........(1)

where,

q = heat absorbed

m = mass of substance

$c$ = specific heat capacity of substance

$\Delta T$ = change in temperature

To calculate the amount of heat required at same temperature, we use the equation:

$q=m\times \Delta H$ ........(2)

where,

q = heat absorbed

m = mass of substance

$\Delta H$ = enthalpy of the reaction

The processes involved in the given problem are:

$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

For process 1:

We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

Putting values in equation 1, we get:

$q_1=42.0g\times 2.3J/g.K\times 43K\\\\q_1=4153.8J$

For process 2:

We are given:

Conversion factor: 1 kJ = 1000 J

Molar mass of ethanol = 46 g/mol

$m=42.0g\\\Delta H_{vap}=38.56kJ/mol=\frac{35.56kJ}{1mol}\times (\frac{1000J}{1kJ})\times (\frac{1}{46g/mol})=773.04J/g$

Putting values in equation 2, we get:

$q_2=42.0g\times 773.04J/g\\\\q_2=32467.7J$

Total heat required = $[q_1+q_2]$

Total heat required = $[4153.8J+32467.7J]=36621.5J$

Hence, the total heat required is 36621.5 J

For b:

The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$