Part c suppose 1 kg each of water (4.19 j/(g ⋅ ∘ brick (0.90 j/(g ⋅ ∘ iron (0.46 j/(g ⋅ ∘ and olive oil (1.79 j/(g ⋅ ∘c)) were held at the same initial temperature and heated for an equivalent amount of time. indicate their relative final temperatures from lowest (left) to highest (right). assume no heat is lost to the surroundings. you can use the intro tab of the phet to visualize the temperature changes. (assume that the mas of the water, iron, and olive oil is 1 kg each and that the mass of the brick is 0.5 kg.) rank from lowest resulting temperature to highest resulting temperature. to rank items as equivalent, overlap them. view available hint(s) ironwaterolive oilbrick

Explanation:

The change in temperature may be calculated from the formula:

Where:

Then, you can make these assumptions or inferences from the conditions stated in the problem:

Then, you can state that, for those samples, ΔT = k / C, i.e. the increase in temperature is inversely related to the specific heat.

That means that the higher the specific heat the lower ΔT, and the lower the specific heat the higher ΔT.

The ranking in decrasing order of specific heat is:

Ranking in increasing order of ΔT:

And since all of them started at the same temperature, that is the ranking in resulting temperature from lowest to highest:

That means that the sample of water, the matter with the highest specific heat capacity (4.19 J/g°C), will reach the lowest temperature, and the sample of iron, the matter with the lowest heat capacity (0.46 J/g°C) will reach the highest temperature.

the answer is c i did the same question so i know that's right!

when insulating materials rub against each other, they may become electrically charged . electrons , which are negatively charged, may be 'rubbed off' one material and on to the other. the material that gains electrons becomes negatively charged. the material that loses electrons is left with a positive charge.