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Sonja [21]
3 years ago
12

A substance releases 2,192 J of heat, changing its temperature from 50 degrees C to 5 degrees C. The mass of the substance is 73

.9 grams. What is the specific heat of the substance? Round your answer to TWO digits after the decimal point.
Chemistry
1 answer:
Gre4nikov [31]3 years ago
3 0

The specific heat capacity of the given substance is -0.66 J/g°C.

<u>Explanation:</u>

The heat absorbed by any substance is the product of its mass, specific heat capacity and change in temperature.

q = m × c × ΔT

m is the mass in grams

q = amount of heat released or absorbed in J

ΔT = change in temperature in °C = 5 -50 = -45°C

c = specific heat capacity in J/g°C

c = $\frac{q}{m \times dT}

Plugin the values, we will get,

c = $\frac{2192 }{73.9 \times-45}

= -0.66 J/g°C

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a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp
Snowcat [4.5K]

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

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3 years ago
If a system performs 147 kJ of work while receiving 47 kJ of heat, what is change in its internal energy?
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Answer:

-100 kJ

Explanation:

We can solve this problem by applying the first law of thermodynamics, which states that:

\Delta U = Q-W

where:

\Delta U is the change in internal energy of a system

Q is the heat absorbed/released by the system (it is positive if absorbed by the system, negative if released by the system)

W is the work done by the system (it is positive if done by the system, negative if done on the system)

For the system in this problem we have:

W = +147 kJ is the work done by the system

Q = +47 kJ is the heat absorbed by the system

So , its change in internal energy is:

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6 0
3 years ago
The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The
dsp73

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

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