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Ilya [14]
3 years ago
13

The specific heat of a certain type of cooking oil is 1.75 cal/ (g c) how much heat energy is needed to raise the temperature of

2.67 kg of this oil from 23 c to 191 c
Chemistry
1 answer:
sineoko [7]3 years ago
4 0

Just use the formula:


Q = cMΔT


where

Q = heat energy needed for that material to get desired temperature change (in Joules)

M = mass (in grams) so you have to convert from kilograms.

c = specific heat constant for the material being heated [in /(grams oC)]

ΔT = change in temperature (in oC)

Q = (1.75)(2.17 x 1000)(191 - 23) = (1.75)(2170)(168) = 637,980 Joules

hope that helps you in some way

also have a great day!

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Answer:

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The average heat evolved by the oxidation of foodstuffs is then:

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the heat produced after 3 h would be:

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Assume the heat capacity of the body is 4.18 J g⁻¹K⁻¹

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6 0
3 years ago
Duncan takes a break from studying and goes to the gym to swim laps If swimming burns amount of time ? 85 * 10 ^ 5 cal per hour
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When 6.85×10⁵ cal is converted to kilojoules, the result obtained is 2866.04 KJ

<h3>Data obtained from the question </h3>
  • Energy (cal) = 6.85×10⁵ cal
  • Energy (KJ) =?

<h3>Conversion scale </h3>

1 cal = 0.004184 KJ

<h3>How to convert 6.85×10⁵ cal to kilojoules</h3>

1 cal = 0.004184 KJ

Therefore,

6.85×10⁵ cal = 6.85×10⁵ × 0.004184

6.85×10⁵ cal = 2866.04 KJ

Thus, 6.85×10⁵ cal is equivalent to 2866.04 KJ

Learn more about conversion:

brainly.com/question/2139943

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