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2 years ago

Find the amount of heat energy needed to convert 400 grams of ice at -38°C to steam at 160°C.

1 answer:

7 0

The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)

<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>

Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
Initial temperature (T₁) = –25 °C
Final temperature (T₂) = 0 °
Change in temperature (ΔT) = 0 – (–38) = 38 °C
Specific heat capacity (C) = 2050 J/(kg·°C)
Heat (Q₁) =?

Q = MCΔT

Q₁ = 0.4 × 2050 × 38

Q₁ = 31160 J

<h3>How to determine the heat required to melt the ice at 0 °C</h3>

Mass (m) = 0.4 Kg
Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
Heat (Q₂) =?

Q = mL

Q₂ = 0.4 × 334000

Q₂ = 133600 J

<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>

Mass (M) = 0.4 Kg
Initial temperature (T₁) = 0 °C
Final temperature (T₂) = 100 °C
Change in temperature (ΔT) = 100 – 0 = 100 °C
Specific heat capacity (C) = 4180 J/(kg·°C)
Heat (Q₃) =?

Q = MCΔT

Q₃ = 0.4 × 4180 × 100

Q₃ = 167200 J

<h3>How to determine the heat required to vaporize the water at 100 °C</h3>

Mass (m) = 0.4 Kg
Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
Heat (Q₄) =?

Q = mHv

Q₄ = 0.4 × 2260000

Q₄ = 904000 J

<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>

Mass (M) = 0.4 Kg
Initial temperature (T₁) = 100 °C
Final temperature (T₂) = 160 °C
Change in temperature (ΔT) = 160 – 100 = 60 °C
Specific heat capacity (C) = 1996 J/(kg·°C)
Heat (Q₅) =?

Q = MCΔT

Q₅ = 0.4 × 1996 × 60

Q₅ = 47904 J

<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>

Heat for –38 °C to 0°C (Q₁) = 31160 J
Heat for melting (Q₂) = 133600 J
Heat for 0 °C to 100 °C (Q₃) = 167200 J
Heat for vaporization (Q₄) = 904000 J
Heat for 100 °C to 160 °C (Q₅) = 47904 J
Heat for –38 °C to 160 °C (Qₜ) =?

Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Qₜ = 31160 + 133600 + 167200 + 904000 + 47904

Qₜ = 1.28×10⁶ J

Learn more about heat transfer:

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A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

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the great size or extent of something

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A chemist heats 100.0 g of FeSO4 x 7H2O in a crucible to drive off the water. If all the water is driven off, what is the mass o

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FeSO₄*7H₂O(s) = FeSO₄(s) + 7H₂O(g)

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M(FeSO₄)=151.9 g/mol

m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)=m(FeSO₄)/M(FeSO₄)

m(FeSO₄)=M(FeSO₄)m(FeSO₄*7H₂O)/M(FeSO₄*7H₂O)

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Which trend is observed as the first four elements in group 17 on the periodic table are considered in order of increasing atomi

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ANSWER:

The melting and boiling points increase in order of increasing atomic number.

The size of the nucleus increases in order of increasing atomic number.

Ionization energy decreases in order of increasing atomic number.

Electronegativity decreases in order of increasing atomic number.

Electron Affinity decreases in order of increasing atomic number.

The reactivities decrease in order of increasing atomic number.

EXPLANATION:

NAME MELTING POINT BOILING POINT

Fluorine -220 -188

Chlorine -101 -35

Bromine -7.2 58.8

Iodine 114 184

Melting and Boiling points increase as shown above.

NAME COVALENT RADIUS IONIC RADIUS

Fluorine 71 133

Chlorine 99 181

Bromine 114 196

Iodine 133 220

Size increases as shown above.

NAME FIRST IONIZATION ENERGY

Fluorine 1681

Chlorine 1251

Bromine 1140

Iodine 1008

Ionization energy decreases as shown above.

NAME ELECTRONEGATIVITY

Fluorine 4

Chlorine 3

Bromine 2.8

Iodine 2.5

Electronegativity decreases as shown above.

NAME ELECTRON AFFINITY

Fluorine -328.0

Chlorine -349.0

Bromine -324.6

Iodine -295.2

Electron affinity decreases as shown above.

REACTIVITY

The reactivities of the halogens decrease. This is due to the fact that atomic radius increases in proportion with an increase of electronic energy levels. This decreases the pull for valence electrons of other atoms, minimizing reactivity.

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