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3 years ago

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Specific Heat Material Copper Aluminum Gold Pyrex Specific Heat (J/g °C) 0.385 0.902 0.129 0.75 Material Water Ice Olive Oil Air

Specific Heat (1/g °C) 4.18 2.01 2.38 1.00 If the same amount of heat was applied to a vessel containing 100.0 g of water and 100.0 g of Olive Oil, which would get hotter? Why? If you had a 1.0 kg pot made of each metal, which pot would allow you to heat water to boiling more quickly? Why? How much heat would it take to raise the temperature of 100.0 g of water from 20.0°C to 100.0 °C? (Ignore the pot)

1 answer:

zimovet [89]3 years ago

3 0

Explanation:

It is known that the amount of heat necessary to raise the temperature of 1 gram of a substance by $1^{o}C$ is known as specific heat.

Since, q = $mC \Delta T$

So, larger is the specific heat of a substance less will be the change in its temperature.

Therefore, olive oil has less specific heat as compared to water. This means that olive oil would get hotter.

Similarly, the specific heat of gold is lesser than the given materials or metals. Hence, gold will requires less heat to rise its temperature.

As a result, water present in gold will heat readily.

As the relation between heat and specific heat is as follows.

q = $mC \Delta T$

Therefore, calculate the amount of heat required by the water as follows.

q = $mC \Delta T$

= $100 g \times 4.18 J/g^{o}C \times (100 - 20)^{o}C$

= 33440 J

or, = 33.44 kJ (as 1 kJ = 1000 J)

Thus, 33.44 kJ heat would it take to raise the temperature of 100.0 g of water from $20.0^{o}C$ to $100.0^{o}C$ .

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