Answer:
[tex]c=0.45\ J/g^{\circ} C[/tex]
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,
[tex]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\times (c)\times (40.22-800.00)\\\\8552.096 = 18994.5c\\\\c=\dfrac{8552.096 }{18994.5}\\\\c=0.45\ J/g^{\circ} C[/tex]
So, the specific heat of iron is [tex]0.45\ J/g^{\circ} C[/tex]
