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 °c to increase its temperature from 25 °c to its melting temperature of 1535 °c?

Answer 1

`Q=50,849.25J\\\\Q=12,153.3cal`

Step-by-step explanation:

Hello!

In this case, given the mass, temperature change and specific heat, it is possible to compute the required heat in joules as shown below:

`Q=mC(T_2-T_1)\\\\Q=75.0g*0.449(J)/(g\°C)(1535\°C-25\°C)\\\\Q=50,849.25J`

Now, since 1 cal =4.184 J, this result in calories is:

`Q=50,849.25J*(1cal)/(4.184J)\\\\Q=12,153.3cal`

Best regards!