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eimsori [14]
3 years ago
10

Which of the following substances would show the smallest temperature change upon absorbing 2.35 kJ of heat? (MTS 2/10/2018) 27

g Fe ; cFe = 0.449 J/goC 27 g H2O ; cH2O = 4.18 J/goC 27 g ethanol ; cethanol = 2.42 J/goC
Chemistry
1 answer:
Darina [25.2K]3 years ago
7 0

Answer:

Water.

Explanation:

Hello,

Based on the formula:

Q=mCpΔT

Solving for ΔT, we state that:

Δ[tex]T=\frac{Q}{mCp}[/tex]

Now, for each substance we get:

ΔT_{Fe}=\frac{2350J}{27g*0.449J/g^0C} =193.85^0C\\

ΔT_{H_2O}=\frac{2350J}{27g*4.18J/g^0C} =20.82^0C\\

ΔT_{EtOH}=\frac{2350J}{27g*2.42J/g^0C} =35.97^0C\\

We sum up that the water shows the smallest temperature change. This could be also stated by know that the iron has the highest heat capacity.

Best regards.

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<span>133.7 mL KOH</span>

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If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did
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<u>Answer:</u>

<em>4.5 L water we have in litres (L).</em>

<em><u></u></em>

<u>Explanation:</u>

Q=m\times c \times \Delta T

where

\Delta T = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0  cal/(g℃))  

m is the mass of water

Plugging in the values  

\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$

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\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$

=4.5 L (Answer)

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