Question

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)

Answer 1

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 $100.0^{o}C$.