Question

How much heat, in joules and in calories, must be added to a 75.0- g iron block with a specific heat of 0.449 j/g degree cecius to increas its temperature from 25 degree celcius to its melting temperature of 1535 dgree celcius?

Answer 1

Answer: The heat required by iron is 50849.25 Joules or 12152.97 Cal.

Explanation:

To calculate the amount of heat released or absorbed, we use the equation:

$q=mc\Delta T$

where,

q = heat absorbed

m = mass of iron = 75 g

c = specific heat capacity of iron = 0.449 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=1535^oC-25^oC=1510^oC$

Putting values in above equation, we get:

$q=75g\times 0.449J/g.^oC\times 1510^oC\\\\q=50849.25J$

Converting the value from joules to calories, we use the conversion factor:

1 J = 0.239 Cal

So, $50849.25J=50849.25\times 0.239=12152.97Cal$

Hence, the heat required by iron is 50849.25 Joules or 12152.97 Cal.