How much heat, in joules and in calories, must be added to a 75.0Â-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?

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