An 89.0-g piece of iron whose temperature is 78.3 degrees C (351.5 K) is placed

Question

An 89.0-g piece of iron whose temperature is 78.3 degrees C (351.5 K) is placed in a beaker containing 240 g of water at 17.0 degrees C (290.2 K). When thermal equilibrium is reached, what is the final temperature? (Assume no energy is lost to warm the beaker and its surroundings.) The specific heat of iron is 0.449 J/g K, and the specific heat of water is 4.184 J/g K. ***Please write the step-by-step. I'm especially confused about how to calculate the "T" on each side of the equation (I don't know how to end up with only one "T" for the left side of the calculation). Thanks!

Explanation / Answer

In the above reaction the amount of heat gained by water = amount of heat lost by iron.

Let us calculate the amount of heat gained by water in this reaction

We know that

The amount of heat energy (q) gained or lost by a substance is equal to the mass of the substance (m) multiplied by its specific heat capacity (Cg) multiplied by the change in temperature (final temperature - initial temperature)

q = m x Cg x (Tf - Ti)

Let us assume Final temperature as Tf

Temperature difference Tf -Ti = (Tf-290.2 ) K and

The specific heat of water is 4.184 joule/gram K

q = 240 g x 4.184 joule/gram K x (Tf-290.2 ) K

   =  1004.16x(Tf-290.2 ) Joules

Hence the heat energy lost by the metal = -1004.16x(Tf-290.2 ) Joules

q = m x Cg x (Tf - Ti)

-1004.16x(Tf-290.2 ) Joules = 89.0 g x 0.449 J/g K x (Tf-351.5) K

                                             = 39.961 x (Tf-351.5) Joules

If solved the above equation we will get Final temperature Tf = 292.55 K ( 19.55 C)

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