After reading the answers and many of the comments, I realised I did not put nearly enough restrictions on this question. The scenario I am assuming is as follows:

There are two cups of water of identical volume in cups that do not affect heat exchange (i.e. only the air temperature matters). One is at, say 80 C, and the other is at 5 C. The ambient air temperature is 25 C.

Obviously the warmer cup of water will cool faster than the colder cup, but my question is exactly why?

Again, I think it has to do with Newton's Law of Cooling and not much else. A friend of mine thinks it has more to do with thermodynamics, such as specific heat capacity varying at different temperatures.

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    $\begingroup$ Can you give a concrete example of this happening? A(n extreme) counter example is that if I throw a glass of water on the Sun, I expect the water to heat up pretty substantially pretty fast, whilst I do not expect the temperature of the Sun to change much at all. $\endgroup$ – jacob1729 Jul 17 '19 at 19:40
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    $\begingroup$ It's all about temperature difference - not the direction of the difference. $\endgroup$ – user207455 Jul 17 '19 at 19:48
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    $\begingroup$ This definitely isn't true in general. $\endgroup$ – JMac Jul 17 '19 at 19:51
  • $\begingroup$ For ice the phase transition en.wikipedia.org/wiki/State_of_matter#Phase_transitions needs energy transfer without really change in temperature. $\endgroup$ – HolgerFiedler Jul 18 '19 at 4:20

Temperature difference is one thing. Newtons law of cooling will give that. However; there are many different methods for cooling and heating. In addition to conduction there is also heat transfer from convection (fluids) and radiation. Materials also have a thermal conductivity meaning different materials heat and cool differently. This conductivity is also temperature dependent; most notable at phase transitions (eg compare the conductivity of ice/water/steam - ice has higher thermal conductivity so this would counter your statement, but then there is also no convection, which is why cold water can feel colder than ice).

In a low radiant environment an object will generally have a net heat loss from radiation making the cold object warm slower than a warm onject cools. On a cold sunny day, the statement won't be true. Cold things will warm faster than hot things cool.

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Your question has no straightforward answer. How is the temperature being changed and what is being changed? Is it by contact? Conduction? Energy input?

Take, for example, Newton's Law of Cooling (an empirical law or better phrased, some first-order approximation). In this scenario, the heat flow and hence the rate of temperature change is directly proportional to the difference in temperature so heating and cooling occur at the same rate.

If you are talking about temperature gradients, say in a gas, i.e. how fast heat flows through, this depends on another quantity $\kappa$ known as the thermal conductivity. This value may depend on many many parameters.

If you are talking about giving some thermodynamic system energy at a constant rate and watching the temperature change, this is dependent on the heat capacity $C$ which may be higher/lower at different temperatures. It also depends on what other quantities are being held constant.

Altogether, your claim is

  1. unclear given the context you ask it in or
  2. if it is clear, untrue in general

There are many more contexts you could think about but especially in thermodynamic problems, you need to be very clear on what is being held constant/what process is exactly taking place to get a meaningful answer. I simply wanted to demonstrate the ambiguity the question presents.

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The rate at which the temperature of a solid object of mass $m$ (kg) and specific heat capacity $\tilde{C}_p$ (J/kg $^o$C) will change $dT/dt$ ($^o$C/s) is written as below.

$$ m \tilde{C}_p\ {dT}/{dt} = \dot{q} $$

The term $\dot{q}$ (J/s) is the rate at which heat enters (to heat up) or leaves (to cool down) the object.

To start, assume the object is initially at $T_o$ and the surroundings is held at constant $T_s$.

Types of Heat Flow

Heat can enter/leave the object by radiation to a surrounding fluid (gas or liquid). The radiation rate is proportional to the fourth power of absolute temperature $\dot{q}_r \propto T^4$. The initial overall rate of heat transfer is expressed to first order as below.

$$\dot{q}_{rT} \propto |T_o^4 - T_s^4|$$

You can use this to determine the initial temperatures of the object where the cold object receives more radiation heat flow from the surroundings than the hot object looses by radiation heat flow to the surroundings. Thus, you can determine for which cases a cold or hot object changes temperature faster than a hot object in the same air temperatures. Note that a vacuum has a temperature $T_a = 0$ K and therefore has not radiation input to the object. In this case, the hot and "cold" object will both cool down.

Heat can enter/leave the object by convection when the surroundings is a fluid (gas or liquid). The rate of heat transfer is proportional to temperature difference. Using the same process as above, you will find the net rate of heat transfer by convection is determined as below.

$$ \dot{q}_{hT} \propto |T_o - T_s| $$

We learn from this that a hot or cold object will cool or heat by convection at the same rate as long as the temperature difference between the object and the surrounding fluid is the same.

Finally, heat can enter/leave the object to the surroundings by conduction in the surroundings. This happens with a temperature gradient $dT_s/dz$ ($^o$C/m) in the surroundings. A fluid does not typically sustain a temperature gradient well; the gradient dissipates and the flow is by convection. Therefore making a statement that you want to consider conduction in the surroundings is akin to saying that you define the surroundings also as a solid. The net rate of conductive flow is written as below.

$$ \dot{q}_{kT} \propto |{dT_s}/{dz}| $$

Combined Analysis

The final equation will have a form as below.

$$ m \tilde{C}_p\ {dT_o}/{dt} = \sigma A \left( \epsilon_s T_s^4 - \epsilon_o T_o^4 \right) + h_s A \left(T_s - T_o\right) - k_s A\ {dT_s}/{dz} $$

The various terms include the Stefan-Boltzmann factor $\sigma$, object area $A$, convection coefficient $h_s$, and thermal conductivity $k_s$. This equation defines the rate of temperature change of the object at any point in time. The +/- signs are set so that, when the object is hotter than the surroundings, the object will cool down (and vice-versa). It is a rather unwieldy equation that is typically solved by considering only one of the three cases (radiation, convection, or conduction) at a time. It also presumes the object heats or cools uniformly throughout. The case when you have a thermal gradient in the solid object is yet another equation.


As a general rule, the rate at which a hot object changes temperature (cools down) cannot be stated to be faster (or slower) than the rate at which a cold object changes temperature (heats up). The rate of heat transfer depends on the type of surroundings (vacuum, fluid, or solid), the temperature of the surroundings, and the type of heat transfer (radiation, convection, or conduction).

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