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

Water < Olive oil < Brick < Iron

Explanation:

The change in temperature may be calculated from the formula:

Q = m × C × ΔT ⇒ ΔT = Q / (m × C)

Where:

Q = amount of heat energy supplied (or released in case of cooling)m = massC = specific heat (a different constant, property, for every matter)ΔT = increase of temperatute (decrease when it is cooling)

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

Q is the same for all the matters because it is supplied by a external source, taking into account that all the substances are held the same time.Initial temperature is the same for all the substances (given)The mass is the same (1 kg each sample)

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:

Water (4.19) > Olive oil (1.79) > Brick (0.9) > Iron (0.46)

Ranking in increasing order of ΔT:

Water < Olive oil < Brick < Iron

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

Water < Olive oil < Brick < Iron ← answer

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.

Approximately 11.43

Explanation:

Using the formula charge=current×time

then current=charge/time. That then will be C=96/8.4=11.4285714286 so approximately 11.43