In one of my experiment, I wrapped aluminium wire around the Cu hollow tube and applied some voltage across the Al wire. So the Cu tube got heated up. My question is how to know that how much temperature rise of Cu tube will take place for a given power at a certain time theoretically, so that I can match it with the thermocouple temperature that I got.
Measure resistance of Cu tube, apply Joule's heating law, you will get theoretical power of heat generated in the Tube.
=> "Power of heating generated by an electrical conductor is proportional to the product of its resistance and the square of the current: P ∝ I2R " [ref: https://en.wikipedia.org/wiki/Joule_heating]
The energy dissipated by resistance heating would be given by: V·I·t = I2·R·t, where V stands for voltage across the resistor, I is the electric current, R is resistance, and t is time. Nearly, all the supplied electric energy would be converted to heat. Part of that heat can in be expected to be lost to the surroundings. The remaining heat increases the temperature of the resistor (Tr) and of the tube being heated (Tt ≤ Tr). When the resistor temperature becomes sufficiently high, the heat transfer between the resistor and the tube can be chiefly attributed to radiation. Seemingly, for a first simplified approach, black body conditions can be judged acceptable. As long as thermal equilibrium can be nearly attained, what could require some thermal insulation from surroundings, we approximately should have: I2·R ≈ A·σ·(Tr4 - Tt4) ≈ rate of heat loss to surroundings, where A stands for the effective radiating area and σ is the Stefan–Boltzmann constant. It was assumed, for simplicity, that either no direct contact occurs between the resistor and the tube, or that contact area is rather small. Otherwise, the heat conduction between the resistor and the tube could not be (reasonably) neglected. For low enough temperatures, the heat conduction contribution would become the most significant. We have accepted that most of the heat generated by the resistor is first-hand transmitted to the tube, and not directly to surroundings, from which the resistor is thought to be thermally insulated to a reasonable extent.
Basically, it is extremely difficult to calculate temperatures. Engineers use mostly simulations or finite element analysis. Convection of surrounding air is also important factor to be considered.
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