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?

1 Answer
Anjali G
May 9, 2018

#Q = 50,800 " J" = 12,100 " cal"#

Explanation:

Use the equation that relates mass, specific heat, and change in temperature to the amount of heat of a substance:
#Q = mc Delta t#

*Note that this equation assumes that temperature has units of Kelvin. We do not need to convert from Celsius to Kelvin because change in Celsius is the same as change in Kelvin.

#Q = (75.0 g) (0.449 frac{J}{g* ^oC})(1535^oC - 25^oC)#

#Q = 50,849.25 " J" -> "(Significant figures) " Q = 50,800 " J"#

#Q = 50,849.25 " J" *frac{1 " calorie"}{4.186" J"} = 12,147.46 " cal"#

#-> "(significant figures) " Q = 12,100 " cal"#

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