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
