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

Note that
1 cal = 4.184 J

Given:
m = 75.0 g, the mass of the iron block
c = 0.449 J/(g-°C), the specific heat of iron
Tm = 1535 °C, melting temperature
T0 = 25 °C, initial temperature.

Sensible heat required to raise the temperature of the iron is given by
Q = mc(Tm - T0)
    = (75 g)*(0.449 J/(g-°C))*(1535 - 25 °C)
    = 50849.25 J = 50.849 kJ

Q = (50849.25 J)*( 1 cal/4.184 J)
    = 12153.3 cal

Answer:
50849.3 J or 12153.3 cal