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

50713.5 joules or 12121.78 calories of heat must be added to a 75.0-g iron block to increase its temperature from 25 °C to 1535 °C, with a specific heat of 0.449 J/g°C.

Step-by-step explanation:

To calculate how much heat must be added to a 75.0-g iron block to increase its temperature from 25 °C to its melting temperature of 1535 °C, we use the formula Q = mcΔT, where Q is the heat in joules, m is the mass, c is the specific heat, and ΔT is the change in temperature.

The specific heat of iron is given as 0.449 J/g°C, the mass m is 75.0 g, and the change in temperature ΔT is (1535 - 25)°C = 1510°C.

Using the formula:

Q = (75.0 g) × (0.449 J/g°C) × (1510°C)

Q = 50713.5 joules

To convert joules to calories, we use the conversion 1 calorie = 4.184 joules:

Q = 50713.5 joules × (1 calorie / 4.184 joules)

Q = 12121.78 calories

Hence, 50713.5 joules or 12121.78 calories of heat must be added to the iron block.

Answer 2

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