A solid copper cube with sides 50 mm long is heated in a furnace to a uniform of 450 °C. The cube is than removed from the furnace and placed on a flat surface of low thermal conductivity, so that the heat transfer from the bottom face of the cube can be neglected, and allowed to cool in air at 25 °C. Calculate the time required for the temperature of the cube to fall to 300 °C. Take the thermal conductivity, specific heat capacity and density to be 390 W/(m.K), 410 J/(kg.K) and 7500 kg/m^3 respectively and the surface heat-transfer coefficient to be 5 W/(m^2.K) for the five exposed faces of the cube. Anye help would be greatly appreciated.
Hello I think this is an application of Newton's Cooling law You must have some formulas / differential equations / principles to apply if you're doing a physics course. You've got a lump of hot copper with 5 radiating surfaces of known area that is sitting in are room of constant temperature . You have some constants that allow you to work out its thermal mass, and the rate at which it dissipates energy and need to know how long it will take to drop from T1 to T2. The cooling curve should look like Ae^-kt derived from some differential equations. Hope this helps as a start J
You have some constants that allow you to work out its thermal mass, and the rate at which it dissipates energy and need to know how long it will take to drop from T1 to T2.