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laila [671]
3 years ago
9

How much energy does it take to melt a 16.87 g ice cube? ΔHfus = 6.02 kJ/mol How much energy does it take to melt a 16.87 g ice

cube? = 6.02 kJ/mol 108 kJ 102 kJ 5.64 kJ 936 J none of the above
Physics
1 answer:
Vilka [71]3 years ago
5 0

Answer:

How much energy does it take to melt a 16.87 g ice cube? ΔHfus = 6.02 kJ/mol How much energy does it take to melt a 16.87 g ice cube? = 6.02 kJ/mol

A. 108 kJ

B. 102 kJ

C. 5.64 kJ

D. 936 kJ

E. none of the above

<em>5.64 kJ</em>

Explanation:

The Heat of fusion is the heat energy required to dissolve a given mass of ice at melting point.

<h3>Step by Step Calculation</h3>

The heat energy required to dissolve ice can be calculated using the expression below;

Q = ΔH_{f} x m ...............................................1

where Q is the heat energy required;

           ΔH_{f}  is the heat of fusion for ice;

           m is the mole

All the parameters above are provided in the question except m, so to get m we use the molar mass of water (also for ice) which is 18.01528 g/mol .

<em>This means that 18.01528 g of ice is contained in one mole, therefore the mole for 16.87 g of ice is given as;</em>

m = \frac{16.87g}{18.015g/mol}

m = 0.9364 mole of ices

Now the parameters are complete, we are given;

ΔH_{f}  = 6.02 kJ/mol

m = 0.9364 mol

Q =?

Substituting into equation 1, we have

Q =  6.02 kJ/mol x 0.9364 mol

Q = 5.64 kJ

<em>Therefore, the energy required to melt 16.87 g of ice is 5.64 kJ</em>

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\texttt{ }

<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

\boxed{E_p = \frac{1}{2}k x^2}

where:

<em>Ep = elastic potential energy ( J )</em>

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Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

F = k x

mg = k x

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k = 445 \frac{5}{11} \texttt{ N/m}

\texttt{ }

<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

Ep_1 + Ek_1 = Ep_2 + Ek_2

\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek

Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2

Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh

Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)

\boxed {Ek \approx 1.20 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Young Modulus : brainly.com/question/9202964
  • Simple Harmonic Motion : brainly.com/question/12069840

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

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