Transfer specific heat of a metal to water - comparing metals

Question

Specific Heat of Gold: 130J / kg K, Iron 473 J / kg K

Assume both of them have equal mass and are heated in boiling water.

Then each sample is dropped into a different beaker of water. All of the beakers contain same volume of water at the same temperature.

Which beaker will have the lowest final water temperature?

How to express it in math to proof which one will have the lower final water temperature.

q (metal) = - q (water) is this correction?

Because of the " -" sign is on the q (water) side, it seems the bigger minus has the lower water temperature.

But I think the one with the lower specific should have the lower final water temperature.

How to proof from the equation?

Answer 1

Assuming the new beakers of water into which the metals are dropped is at a temperature below boiling, say at room temperature for example, then -q(metal) = +q(water).

The metal with the greater specific heat (iron) will have more joules of energy than the gold when both metals are at the same temperature. They will then lose that heat to the water so the one with the lower specific heat (gold) will produce a lower temperature of the water than will the iron.

Proof from equation:

q = mC∆T and solve for ∆T

Since the mass (m) and the temperature are the same (constant), q will vary only with the specific heat (C).

q = C x a constant

The metal with the lowest C will possess the least amount of heat and hence that sample of water will have a lower final temperature.