Using only a 2-kilowatt heater (that runs at 80% efficiency), you heat a 1-kilogram piece of copper. Plot the temperature vs time graph of copper over a period of time that you consider relevant. For the latent heat of fusion of copper use 200 kJ/kg.Not exactly sure where to start with this problem. i know what a heating curve is and what latent heat of fusion is, but i just don't know how to piece it all together for this problem.Thanks so much for all your help!!!
let's make the assumption that the CU block starts around room temperature and call that Troom = 20?C Heat is supplied to the block at a rate of 2 kJ/sec (2 kW). The 80% efficiency is irrelevant: the heater supplies 2 kW to the block regardless of what input power it needs. The block starts heating at a rate ?Q/?t = mCv?T/?t ?Q/?t = 2x10^3 J/s m = 1kg Cv = specific heat of Cu (look it up) ?T/?t = temperature rise rate. Since ?Q/?t is constant, ?T/?t is constant, so the temperature will rise at a rate of ?T/?t =[?Q/?t]/mCv It will continue rising until it gets to the melting temperature, Tm, of Cu (look it up). How long does this take starting from 20?C? At this point, the temperature will stop rising until the whole Cu block is molten. Call H the latent heat of fusion. It will take mH Joules to melt the block. Energy is supplied at a constant ?Q/?t rate, so mH Joules will be supplied in [?Q/?t]tmelting = mH. Solve for tmelting to get how long it takes to melt. After that the temperature will keep rising until the the block reaches the boiling point of Cu .... You should be able to figure out what happens next.
from what I read you can expect the copper to gain 1.6 kJ/s so E = 1.6 t probably OK to take To as 20 C - room temp then E = m c delta(T) delta(T) = E/ (m c ) look up c - specific heat of solid copper - get it in kJ / (kg*C) so temp as a function of time as the copper warms to its melting point T = To + delta(T) = To + E/(m c) = To + 1.6 t / (m c) use this to generate temps from time 0 up to the melting T then you will have a flat portion as the copper melts without energy change use E = m Lf to get total needed to melt it and E = 1.6 t to get the time for the graph finally look up c for liquid copper and use T = To + 1.6 t / (m c) with To as the MP of copper maybe run the time beyond melting for about the same amount of time it took to warm the solid copper to MP - this would give some symmetry to your graph you would wind up with a graph having two upward sloping segments joined by the horizontal melting event
