Question

500 grams of an unknown substance gains 23000 J of heat as it is heated from a temperature of 250C to 1250C. Using mathematical computation and the chart below, determine the identity of the unknown substance.

Answer 1

Answer:

The given substance is cast iron.

Explanation:

Given data:

Mass of substance = 50 g

Heat absorbed = 23000 J

Initial temperature = 250°C

Final temperature = 1250°C

Which metal is this = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 1250°C -  250°C

ΔT = 1000°C

23000 j = 50 g ×c ×1000 °C

23000 J = 50,000 g. °C×c

c = 23000 J /50,000 g. °C

c = 0.46 J/g.°C

The given substance is cast iron.