The liquid in one container drops 100 F, while the same liquid in a different container drops 100C. How does the change in thermal energy in the two compare? Explain your answer.

The thermal energy is given as

E=mc (T2-T1)

For the temperature change.

ΔT can be expressed in units of Kelvin or degrees Celsius. The ΔT of Kelvin is equal to that of ΔT of Celsius but not equal to the ΔT of Fahrenheit.

Therefore change in temperature of 100F is not equal to change in temperature of 100C

°F = 9/5 °C + 32

So let assume 20°C, so the increase of 100°C will give 120°C.

Then °F = 68°F

Now the equivalent of 20°C is 68°F.

So let see the value of 120°C, which is the increase given.

°F = 9/5 * 120+32

°F = 248°F

The change in temp in Fahrenheit is 248-68=180°F

1°F change is = to 0.56°C change

Therefore, 100°F change in temperature in Fahrenheit is equal to 56°C in Celsius

Therefore the thermal energy is not the same, since they have are the same mass and specific heat but different changer in temperature.

So using the formulae given above, the thermal energy of the 100F change is greater than that of the 100C change ...