A51.9g 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

i think its 100

i hope this !