Cooling/heating temperature rate change

[Ihaveaquestion]

Ok, so I could do an experiment on this first, but I decided to see if anyone knew beforehand. So basically my question is will different properties of the same temperature, exposed to the same temperature, cool/heat at the same rate? For example would a more viscous liquid cool as fast as less viscous liquid of the same temperature undergoing the same cooling? I could use other examples but I think you get the gist. Also, I would appreciate explanations instead of just "yes" or "no". I know my question is very broad, but I appreciate any replies. Thanks.

 

[Drakkith]

That all depends on the heat capacity and the thermal conductivity of the material. In general, the more degrees of freedom the molecules making up the substance have, the greater the heat capacity. See here for more info: https://en.wikipedia.org/wiki/Heat_capacity#Factors_that_affect_specific_heat_capacity

Thermal conductivity is much harder to explain. At least for me. Try this article: https://en.wikipedia.org/wiki/Thermal_conductivity#Influencing_factors

 

[Ihaveaquestion]

That all depends on the heat capacity and the thermal conductivity of the material. In general, the more degrees of freedom the molecules making up the substance have, the greater the heat capacity. See here for more info: https://en.wikipedia.org/wiki/Heat_capacity#Factors_that_affect_specific_heat_capacity

Thermal conductivity is much harder to explain. At least for me. Try this article: https://en.wikipedia.org/wiki/Thermal_conductivity#Influencing_factors

I am still confused :/ I should mention that I have no knowledge in physics and very minimal in math. I am not super smart or anything either, just a very curious person lol.

 

[Drakkith]

Unfortunately I don't think I'll be able to explain any better that those links. Perhaps someone else here can.

 

[Ihaveaquestion]

Unfortunately I don't think I'll be able to explain any better that those links. Perhaps someone else here can.

All good. I appreciate you trying to explain to me.