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.

Step-by-step 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* 0.449J/g.^oC* 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* 0.239=12152.97Cal`

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