Chemistry College

The specific heat capacity of benzene (C6H6) is 1.74 J/g*K. How much energy as heat is required to raise the temperature of 50.00 mL of benzene from 25.52 C to 28.75 C. Density of benzene is 0.876 g/cm3.

Answers

Answer 1

The amount of heat required is 246.165 Joules.



We can use the following method to figure out how much energy it will take to raise the temperature of benzene:

Q = mcΔT

Where Q = the amount of energy needed (in Joules).

m = the number of grams of benzene.

c = the amount of heat that benzene can hold (in J/g*K).

T = temperature change (in Kelvin)

But since the volume of benzene is given to us, we need to find the mass of Benzene by:


Density = mass/volume

mass = density x volume

mass = 0.876 g/cm3 x 50.00 mL = 43.8 g

Change in temperature = 28.75 - 25.52 = 3.23 K

Now, we can use the given numbers to fill in the formula for Q:

Q = mcΔT

Q = 43.8 g x 1.74 J/g*K x 3.23 K

Q = 246.165 J

So, it takes 238.92 Joules of heat energy to raise the temperature of 50 mL of benzene from 25.52°C to 28.75°C.

To know more about calorimetry:

https://brainly.com/question/24813298

Answer 2

Answer:

[tex]\Large \boxed{\boxed{\textsf{heat required = 246.2 J (4 s.f)}}}[/tex]

Explanation:

The heat energy required to raise the temperature of the reaction, i.e, the amount of heat energy released in the reaction (noting the rise in temperature, this is an exothermic reaction), can be calculated with the following calorimetry equation:

[tex]\Large \boxed{\textsf{$q=mc\Delta T$}} \sf \,, where:\\ \\\bullet q = quantity\,of\,heat\,released\,(measured\,in\,joules)\\ \bullet m = solvent\,of\,mass\,(measured\,in\,g\,or\,kg)\\\bullet c = specific\,heat\,capacity\,of\,solution\\\bullet \Delta T = change\,in\,temperature\,of\,solution[/tex]

To solve this, we know that:

[tex]\textsf{$\bullet$ m = 43.8 g (1 mL = 1 cm$^3$, using density = g/cm$^3$)}\\\textsf{$\bullet$ c = 1.74 J/g/K}\\\textsf{$\bullet \Delta T$ = 3.23 ($\Delta T$ is the same whether Kelvin or Celsius is used)}[/tex]

Inputting these values into the formula:

[tex]\large \textsf{$q = (43.8)(1.74)(3.23)$}\\ \\\large \textsf{$\therefore q=246.2$ J}\\ \\ \\\Large \boxed{\boxed{\textsf{$\therefore$ heat required = 246.2 J (4 s.f)}}}[/tex]

To learn more about energy changes in chemical reactions:

https://brainly.com/question/8303247

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