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Ganezh [65]
3 years ago
8

A sample of an unknown substance has a mass of 0.158 kg. If 2,510.0 J of heat is required to heat the substance from 32.0°C to 6

1.0°C, what is the specific heat of the substance?
0.171 J/(gi°C)
0.548 J/(gi°C)
15.9 J/(gi°C)
86.6 J/(gi°C)
Chemistry
1 answer:
blondinia [14]3 years ago
6 0
Specific heat is the amount of heat absorb or released by a substance to change the temperature to one degree Celsius. To determine the specific heat, we use the expression for the heat absorbed by the system. Heat gained or absorbed in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1)

By substituting the given values, we can calculate for C which is the specific heat of the material.

2510 J = .158 kg ( 1000 g / 1 kg) (C) ( 61.0 - 32.0 °C)
C = 0.5478 J / g °C
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Answer:

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Explanation:

Step 1: Data given

Volume of the flask = 500 mL

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Number of moles CO2 = 0.70 moles

The equilibrium constant K for this reaction is 3.80

Step 2: The balanced equation

CO(g) + H2O(g) ⇆ CO2(g) +H2(g)

Step 3: Calculate the initial concentrations

Concentration = moles / volume

[CO] = 1.0 moles / 0.500 L = 2.0 M

[H2O] = 1.5 moles / 0.500 L = 3.0 M

[CO2] = 0.70 moles / 0.500 L = 1.4 M

[H2] = 0M

Step 4: The concentration at the equilibrium

For 1 mol CO we have 1 mol H2O to produce 1 mol CO2 and 1 mol H2

[CO] = 2.0 -X M

[H2O] =  3.0 - X M

[CO2] =1.4 + X M

[H2] = X M

Step 5: Define Kc

Kc = [CO2][H2]/ [CO][H2O]

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X = 1.38

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[CO2] =1.4 + 1.38 M = 2.78

[H2] = 1.38 M

Kc = (2.78*1.38) / (0.62*1.62)

Kc = 3.8

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Answer:

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