# Heat transfer and temperature difference of 2 liquids

We know that heat transfer is calculated by equation $Q$=$m$×$C_p$×$\Delta$$T. Let's perform 2 experiments- Imagine I have 2 cups with water with same masses(volume). 1 is at 100 degree C and other at .5 degree C. If we pour these 2 cups of water in another bigger cup.how fast it's temperature reaches equilibrium. Same test I will perform with temperature of second cup at 30 degrees. Equation says heat transfer will be higher if temperature difference is high. That means 1st experiment will reach equilibrium faster than second. Is it right? I am completely confused. It really takes less time to cool a cup of tea at 70 degree C than 100 degree C. ## 2 Answers The equation you give:$$ Q = m C_p \Delta T $$just tells us the total amount of heat transferred, and does not tell us anything about the rate at which the heat transfer occurs. To calculate heat flow we have to solve the heat equation. If you do a physics degree this apparently innocent equation will cause you many hours of frustrated head scratching, so in practice we tend to use simple approximations. In everyday life heat flow tends to be well described by Newton's equation:$$ \frac{dQ}{dt} \propto \Delta T $$so, as you suggest, the greater the temperature difference the faster the heat flow. The experiment you describe isn't really a good way of showing this, because if you mix hot and cold liquid in practice the rate of temperature change will be controlled by how fast you do the mixing. Newton's equation would be more useful if you place the two liquids in contact but don't allow them to mix e.g. have a metal (or some other high thermal conductivity) divider between them. In that case you're quite correct that the initial heat flow will be faster at 100°C than at 70°C. However the 100°C system will take longer to cool because the amoutn of heat, Q \propto \Delta T, that needs to be transferred is greater with a 100°C difference. The temperature difference as a function of time will look like:$$ \Delta T(t) = \Delta T_0 e^{-\alpha t}$$where$\Delta T_0$is the initial temperature and$\alpha$is a constant related to the thermal conductivity (large$\alpha$means high thermal conductivity). If I use this equation to graph the cooling for 70°C and 100°C initial temperatures (choosing a random value of$\alpha$) I get: So even though the 100°C difference initially cools faster the 70°C difference always reaches any specified temperature difference before the 100°C does. • The intuitive reason is that even if one goes faster, it still has to get to 70 C, and then it is at the same state it was the other some time before. – Davidmh Apr 3 '14 at 10:32 • The graph tells us, the heat transfer rate due to temperature difference is significantly less.it does transfer fast but by little magnitude.what factor needs to be considered to replace proportionality in newtons equation? Is it due to heat capacity variation due to higher temperature or thermal conductivity variation due to higher temperature? – Nemu Rozario Apr 4 '14 at 2:56 • @John If you are giving equations, it is more intuitive to write it as$exp(-t/\tau)$, with$\tau\$ a timescale, which is a kind of thermal relaxation time. – Bernhard Apr 4 '14 at 5:45
• @nemu: see the Wikipedia article for details, though all they do is use a constant called the heat transfer coefficient that needs to be determined experimentally. For a rigorous treatment you'd need to solve the heat equation. – John Rennie Apr 4 '14 at 11:38

The exact times to equilibrium are difficult to calculate. It depends on the size of the domain, the method of pouring, surroundings, etc.

You are right the with a higher temperature difference, the heat transfer will be higher. However, as you are getting closer to the equilibrium temperature, the heat transfer will also decrease. So it is only the initial heat transfer that is larger.

Another way to look at it. The 100 degree C temperature difference situation, will at some point only have 70 degree C temperature difference. From that point, the situation will be the same as a initial temperature difference of 70 degrees. Therefore, the 100 degree cooling will take longer, by approximately the amount of time it takes to cool to 70 degrees.