A 51.9g sample of iron, which has a specific heat capacity of 0.449·J·g?1°C?1, is put into a calorimeter (see sketch at right) that contains 300.0g of water. The temperature of the water starts off at 19.0°C. When the temperature of the water stops changing it's 20.3°C. The pressure remains constant at 1atm. Calculate the initial temperature of the iron sample. Be sure your answer is rounded to 2 significant digits.

the initial temperature of the iron sample is Ti = 90,36 °C

Explanation:

Assuming the calorimeter has no heat loss to the surroundings:

Q w + Q iron = 0

Also when the T stops changing means an equilibrium has been reached and therefore, in that moment, the temperature of the water is the same that the iron (final temperature of water = final temperature of iron = T)

Assuming Q = m*c * (T - Tir)

mc*cc * (T-Tc) + mir*cir * (T - Tir) = 0

Tir = 20.3 °C + 300 g * 4.186 J/g°C * (20.3 C - 19 °C) / (51.9 g * 0.449 J/g°C)

Tir = 90.36 °C

Note:

- The specific heat capacity of water is assumed 1 cal/g°C = 4.186 J/g°C

- We assume no reaction between iron and water