Calculate the initial temperature of the iron sample.

Question

A 57.2 sample of iron, which has a specific heat capacity of 0.449 J * g^-1 *C^-1 , is put into a calorimeter that contains 150.0 g of water. The temperature of the water starts off at 20.0 C. When the temperature of the water stops changing it's 22.8 C. The pressure remains constant at 1atm.

Calculate the initial temperature of the iron sample. Be sure your answer is rounded to the correct number of significant digits.

I tried to follow the formula but kept on getting the incorrect answer, I'm not sure if I'm doing something incorrectly.

Answer 1

The important concept to grasp in this problem is that the heat LOST by the iron must equal the heat gained by the water (conservation of energy). The equation for heat (gained or lost) is q = mC∆T where

q = heat

m = mass

C = specific heat

∆T = change in temperature

So....

mC∆T for iron = mC∆T for water. So we have the following relationship:

(57.2 g)(0.449 J/gº)(∆T) = (150 g)(4.184 J/gº)(2.8º)

∆T of iron = 68.4º

Since the final temperature of the iron is the same as the final temperature of the water (22.8º), the initial temperature of the iron must have been 22.8 + 68.4 = 91.2ºC

Answer 2

Check this out. The 150 g of water went from 20.0 C to 22.8 C, a change of 2.8 C. To calculate how much energy it took to do this, Use the relation Q = 2.8 C x 4.182 J/º g x 150g = 1756.4 J. This heat energy came from the iron at a higher temperature. To find the initial temperature of the iron, set up the following equation:

1756.4 = 0.449 J/ºg x 57.2g ºC x (T_i - 22.8)ºC multiply the terms in the parenthesis by the product of 0.449 x 57.2 to get 1756.4 = 25.67T_i - 585.6. Solve this equation for T_i. I got 91.2 ºC for the initial temperature of the iron.

Please let me know if this corresponds to the correct answer. Thanks.