Question

How many kilocalories are required to increase the temperature of 15.6 g of iron from 122 °c to 355 °c. the specific heat of iron is 0.450 j/g °c?

Answer 1

Heat require to boil 15.6 g iron from 122 C0to 355 C0 whereas,

$Q = m s dT$

Where, m is mass of iron

s is specific heat of iron

d T is change in temperature in celcius

$= 15.6 g * 0.45 J /g /C * (355 - 122) = 1.63 * 10^3 J$

If  

1 cal = 4.2 J

Then,  

$Q = (1.63 * 10^3) /4.2 = 0.389 K cal$

Thus 0.389 k cal of enrgy  is required by a 15.6 g Fe to reach to 355 C^0