That's not a chemistry problem, it's a physics problem. I say this because I've read it all in Physics but not in Chemistry
And since you asked for explanation I'll try to make it as comprehensive and simple as possible. I'm telling you already that the answer will seem WAY longer than it should be, but 70% of this answer will be explanations, so don't be daunted.
I trust you understand the concept of specific heat (capacity) (e.g, in case of water, 4.184 J/(g*C) tells us that 4.184 Joules of energy is required to raise the temperature of 1 gram of water by 1 C.).
Now, the total QUANTITY, or the AMOUNT of heat supplied is given by the formula:
Q (Total heat energy supplied/taken, whatever the case may be) = m (mass) x c (specific heat capacity) x DIFFERENCE of temperature.
Now the heat that we will supply will raise the temperature of the KETTLE and WATER (The system). We, therefore, have to calculate what amount of energy will first cause the kettle to rise from 18 to 100, then the amount of energy that will cause the water to rise from 18 to 100.
I'll take them individually. Let's start with the kettle:
Amount of energy absorbed by/supplied to the COPPER (KETTLE) = m x c x diff. of temp.
Amount of energy absorbed by/supplied to the COPPER (KETTLE) = 400 x 0.385 x (100-18)
Amount of energy absorbed by/supplied to the COPPER (KETTLE) = 12628 J
Amount of energy absorbed by/supplied to the WATER = 500 x 4.184 x (100-18) = 171544 J
So the above tells us that we will first supply 12628 J of heat to the kettle containing the water. That heat IS enough to raise the temperature of kettle from 18 to 100, but practically that heat will be spread evenly throughout the water and the kettle, so the final temperature of the system (either the water OR the kettle) will not be 100. But when we supply an additional 171544 J of heat, the ENTIRE system's temperature would rise to 100 C, i.e both the water and the kettle would be at 100 C.
So all in all, we are supposed to supply a total of 12628+171544 = 184172 J of heat to raise the temperature of water (and the kettle) from 18 to 100 C.
Now the second part asks for what PERCENTAGE of that heat (i.e 184172 J) is used to heat the COPPER ONLY. That's easy. The formula would be:
Percentage Heat used on COPPER= (Heat supplied to copper) / (TOTAL heat supplied to the SYSTEM, which was Kettle+Water) X 100
Percentage Heat used on COPPER = 12628/184172 X 100
Percentage Heat used on COPPER = 6.86%
The question is WAY shorter than what I've written, but it's the explanations that make up the bulk of this answer. You only need to write the equations in the exam.
