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

The amount of heat needed would be the specific heat multiplied by the mass of the substance and the temperature difference. In this case, the mass would be 75.0–g, the specific heat would be 0.449 j/g °c, and the temperature difference would be 1535 -25= 1510

Then the calculation would be: 0.449 j/g °c * 75g * 1510°c = 50,849.25J

In calorie it would be: 50849.25J / 4.184J/cal= 12,153.26 calorie