Question

6. A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at

30.00°C). The final temperature of the system is 40.22°C. Find the specific heat of iron.
90​

Answer 1

`c=0.45\ J/g^(\circ) C`

Step-by-step explanation:

Given that,

A 25 g sample of iron (initially at 800.00°C) is dropped into 200 g of water (initially at 30.00°C). The final temperature of the system is 40.22°C.

We need to find the specific heat of iron.

It can be calculated as:

Cooler water gains = hot metal loses

mc∆T = - mc∆T

Put all the values,

`200g(4.184\ J/g^(\circ) C)(T_f-T_i) = -25g(c)(T_f-T_i) \\\\200g(4.184 )( 40.22-30.00) = -25* (c)* (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=(8552.096 )/(18994.5)\\\\c=0.45\ J/g^(\circ) C`

So, the specific heat of iron is
`0.45\ J/g^(\circ) C`