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butalik [34]
3 years ago
5

Which of the following processes are spontaneous? Check all that apply.

Physics
1 answer:
-Dominant- [34]3 years ago
5 0

Answer:

(1) A hot drink cooling to room temperature.

(2) The combustion of natural gas.

Explanation:

The spontaneous process is the process in which there is a release of energy and moves towards lower energy and a more thermodynamically stable energy state. All the natural processes are spontaneous.

There are two processes which are spontaneous in the given question are:

(1) A hot drink cooling to room temperature: In this, there is a decrease in energy and also it is a natural process and we know that all the natural processes are spontaneous.

(2) The combustion of natural gas: The fire is an example of an exothermic reaction. The combustion is a combination of a decrease in energy and an increase in entropy. So, this process occurs spontaneously.

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Using kinetic molecular theory and collision theory, explain why the absolute temperature scale (kelvins) is more appropriate th
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A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial tempe
Taya2010 [7]

Complete question:

A 3.53-g lead bullet traveling at 428 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet. J

(b) Find the heat required to melt the bullet. J

Answer:

Part (a) the available kinetic energy of the bullet is 323.32 J

Part (b) the heat required to melt the bullet is 216.17 J

Explanation:

Given;

mass of the bullet = 3.53 g = 0.00353 kg

velocity of the bullet = 428 m/s

initial temperature of the bullet = 40.0°C

final temperature of the bullet =  327°C

specific heat capacity, c= 128 J/(kg · °C)

latent heat of fusion, Hf  = 24.5 kJ/kg

Part (a) the available kinetic energy of the bullet. J

KE = ¹/₂ × mv²

KE = ¹/₂ × 0.00353 × 428²

     = 323.32 J

Part (b) the heat required to melt the bullet. J

This is the thermal energy required to increase the temperature of the bullet and the heat energy required to melt the bullet.

Quantity of heat required to raise the temperature of the bullet:

Q = mcΔT

   = 0.00353 × 128 × (327-40)

   = 0.00353 × 128 × 287

   = 129.68 J

Quantity of heat required to melt the bullet:

Q = mH_f

Q = 0.00353 × 24500

   = 86.49 J

TOTAL energy required to melt the bullet = 129.68 J + 86.49 J

                                                                      = 216.17 J

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Explanation:

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