Answered

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 :

SEBASSANDINGO

Answer:

[tex]Q=50,849.25J\\\\Q=12,153.3cal[/tex]

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:

[tex]Q=mC(T_2-T_1)\\\\Q=75.0g*0.449\frac{J}{g\°C}(1535\°C-25\°C)\\\\Q=50,849.25J[/tex]

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

[tex]Q=50,849.25J*\frac{1cal}{4.184J}\\\\Q=12,153.3cal[/tex]

Best regards!

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