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3 years ago

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How much energy (in KJ) is needed to melt 15g of ice at -4oC and raise it to a temperature of 115oC? Round the closest number, a

nd include units

1 answer:

WARRIOR [948]3 years ago

4 0

Answer:

The energy needed is 45.7659 kJ

Explanation:

The given mass of the ice, m = 15 g

The initial temperature of the ice, T₁ = -4°C

The temperature

The final temperature, T₂ = 115°C

The specific heat capacity of ice, c₁ = 2.09 J/(g·°C)

The latent heat of ice, l₁ = 334 J/g

The heat capacity of water, c₂ = 4.184 J/(g·°C)

The latent heat of vaporization, l₂ = 2260 J/g

The specific heat of steam, c₃ = 2.02 J/g

Therefore, we get the energy needed, ΔQ, as follows;

ΔQ = m·c₁·(T₂ - Tₐ) + m·l₁ + m·c₂(Tₓ - Tₐ) + m·l₂ + m·c₃·(T₂ - Tₓ)

Where;

Tₐ = The melting point temperature of water = 0°C

Tₓ = The boiling point temperature of water = 100°C

∴ ΔQ = 15×2.09×(0 - (-4)) + 15 × 334 + 15×4.184×(100 - 0) + 15×2260 + 15×2.02×(115 - 100) = 45,765.9

The energy needed, ΔQ = 45,765.9 J = 45.7659 kJ.

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Th e molar absorption coeffi cients of tryptophan and tyrosine at 240 nm are 2.00 × 103 dm3 mol−1 cm−1 and 1.12 × 104 dm3 mol−1

nataly862011 [7]

Answer:

5.43x10⁻⁵ dm³mol⁻¹= Concentration of tyrosine

2.58x10⁻⁵ dm³mol⁻¹= Concentration of tryptophan

Explanation:

Lambert-Beer law is:

A = E×C×l

Where A is measured absorbance, E is absorption coefficient, C is concentration of solution and l is optical path length.

For the result at 240nm, it is possible to write:

0.660 = 2.00x10³ dm³mol⁻¹cm⁻¹× Ctry × 1cm + 1.12x10⁴ dm³mol⁻¹cm⁻¹× Ctyr × 1cm <em>(1)</em>

At 280 nm:

0.221 = 5.40x10³ dm³mol⁻¹cm⁻¹× Ctry × 1cm + 1.50x10³ dm³mol⁻¹cm⁻¹× Ctyr × 1cm <em>(2)</em>

<em></em>

Thus:

-2.7× (0.660 = 2.00x10³ × Ctry + 1.12x10⁴ × Ctyr) =

-1.782 = -5.40x10³Ctry - 3.024x10⁴Ctyr

0.221 = 5.40x10³ × Ctry + 1.50x10³ × Ctyr

-------------------------------------------------------------------

-1.561 = -28740×Ctyr

<em>5.43x10⁻⁵ dm³mol⁻¹= Ctyr</em>

Replacing this value in (1) or (2) it is possible to find Ctry, that is:

<em>2.58x10⁻⁵ dm³mol⁻¹= Ctry</em>

4 0

4 years ago

Find the standard enthalpy of formation of ethylene, C2H4(g), given the following data: C2H4(g) + 3 O2(g) --> 2CO2(g) + 2 H2O

Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_2H_4$ will be,

$2C(s)+2H_2(g)\rightarrow C_2H_4(g)$ $\Delta H_{formation}=?$

The intermediate balanced chemical reaction will be,

(1) $C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)$ $\Delta H_1=-1411kJ$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$ $\Delta H_2=-393.5kJ$

(3) $H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)$ $\Delta H_3=-285.8kJ$

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) $2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)$ $\Delta H_1=+1411kJ$

(2) $2C(s)+2O_2(g)\rightarrow 2CO_2(g)$ $\Delta H_2=2\times (-393.5kJ)=-787kJ$

(3) $2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=2\times (-285.8kJ)=-571.6kJ$

The expression for enthalpy of formation of $C_2H_4$ will be,

$\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+1411kJ)+(-787kJ)+(-571.6kJ)$

$\Delta H=52.4kJ$

Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

7 0

3 years ago

An unknown metal has a mass of 86.8 g. When 5040 J of heat are added to the sample, the sample temperature changes by 64.7 ∘ C .

grandymaker [24]

Answer: The specific heat of the unknown metal is $0.897J/g^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $5040$ Joules

m= mass of substance = 86.8 g

c = specific heat capacity = ?

Initial temperature of the water = $T_i$

Final temperature of the water = $T_f$

Change in temperature , $\Delta T=T_f-T_i=(64.7)^0C$

Putting in the values, we get:

$5040=86.8\times c\times 64.7^0C$

$c=0.897J/g^0C$

The specific heat of the unknown metal is $0.897J/g^0C$

4 0

3 years ago

In the laboratory, a student dilutes 13.5 mL of a 11.6 M hydroiodic acid solution to a total volume of 100.0 mL. What is the con

igomit [66]

Answer: The concentration of the diluted solution is 1.566M.

Explanation:

The dilution equation is presented as this: $M_{s} V_{s} =M_{d} V_{d}$ .

·M= molarity (labeled as M)

·V= volume (labeled as L)

·s= stock solution (what you started with)

·d= diluted solution (what you have after)

Now that we know what each part of the formula symbolizes, we can plug in our data.

$11.6M*13.5mL=M_{d} *100mL$

We cannot leave it like this because the volumes must be in Liters, not milliliters. To convert this, we divide the milliliters by 1000.

$13.5mL/1000=0.0135L$ $100mL/1000=0.1L$

Now that we have the conversions, let's plug them into the equation.

$11.6M*0.0135L=M_{d} *0.1L$

The only thing that we need to do now is actually solving the answer.

$M_{d} =\frac{11.6M*0.0135L}{0.1L}$ $M_{d} =1.566M$

From the work shown above, the answer is 1.566M.

I hope this helps!! Pls mark brainliest :)

7 0

2 years ago

vitfil [10]

Suspensions

Explanation:

Suspensions are heterogeneous mixtures that contains large particles that can settle out or be filtered.

Suspensions are mixtures of small insoluble particles of a solid in a liquid or gas.
Examples are:

powdered chalk in water
muddy water
harmattan

The particles in suspension can settle on standing

Learn more:

Suspension brainly.com/question/1557970

heterogeneous mixture brainly.com/question/1446244

#learnwithBrainly

8 0

3 years ago