At very low temperatures, the molar heat capacity of rock salt varies with temperature according to debyeâs t 3 law: c = k t 3 î¸ 3 , where k = 1940 j/(molâ·k) and î¸ = 281 k.
a. weâve learned the relation q = ncât for the molar heat capacity, which works well when the molar heat capacity is a constant. how do you think we should handle a situation in which the heat capacity varies with temperature? (hint: this is what calculus is how did we calculate work by a force that varies with position, like the force from a spring? ) find a way to calculate the amount of heat required to raise the temperature of 1.50 mol of rock salt from 11.0 k to 45.0 k; show your work and explain your reasoning.

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

i think the answer is a and b

i wish you luck