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Yakvenalex [24]
3 years ago
6

Calculate the amount of heat necessary to raise the temperature of 135.0 g of water from 50.4°F to 85.0°F. The specific heat of

water = 4.184 J/g·°C.
Chemistry
1 answer:
MAXImum [283]3 years ago
6 0

Here we have to calculate the heat required to raise the temperature of water from 85.0 ⁰F to 50.4 ⁰F.

10.857 kJ heat will be needed to raise the temperature from 50.4 ⁰F to 85.0 ⁰F

The amount of heat required to raise the temperature can be obtained from the equation H = m×s×(t₂-t₁).

Where H = Heat, s  =specific gravity = 4.184 J/g.⁰C, m = mass = 135.0 g, t₁ (initial temperature) = 50.4 ⁰F or 10.222 ⁰C and t₂ (final temperature) = 85.0⁰F or 29.444 ⁰C.

On plugging the values we get:

H = 135.0 g × 4.184 J/g.⁰C×(29.444 - 10.222) ⁰C

Or, H = 10857.354 J or 10.857 kJ.

Thus 10857.354 J or 10.857 kJ heat will be needed to raise the temperature.

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The vapor pressure of ethanol is 1.00 × 102 mmHg at 34.90°C. What is its vapor pressure at 60.21°C? (ΔHvap for ethanol is 39.3 k
Delicious77 [7]

Answer:

The vapor pressure at 60.21°C is 327 mmHg.

Explanation:

Given the vapor pressure of ethanol at 34.90°C is 102 mmHg.

We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

We have given in the question

P_1=102\ mmHg

T_1=34.90\°\ C=34.90+273.15=308.05\ K\\T_2=60.21\°\ C=60.21+273.15=333.36\ K\\\Delta{vap}H=39.3 kJ/mol

And R is the Universal Gas Constant.

R=0.008 314 kJ/Kmol

ln(\frac{P_2}{102})=\frac{39.3}{0.008314}(\frac{1}{308.05}-\frac{1}{333.36})\\\\ln(\frac{P_2}{102})=4726.967(\frac{333.36-308.05}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{333.36\times308.05})\\\\ln(\frac{P_2}{102})=4726.967(\frac{25.31}{102691.548})\\\\ln(\frac{P_2}{102})=1.165

Taking inverse log both side we get,

\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg

8 0
3 years ago
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