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

the answer is the

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.