Question

When 685 J of thermal energy (heat) is added to 7.9 g of a substance at 31°C, the temperature increases from 31 °C to 98 °C. What is the specific heat of the substance?

Answer 1

Specific heat capacity, = 1.2942 J/g°C

Step-by-step explanation:

Given the following data;

Heat capacity = 685 J

Mass = 7.9 g

Initial temperature = 31°C

Final temperature = 98°C

To find the specific heat capacity of the substance;

Heat capacity is given by the formula;

`Q = mcdt`

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 98 - 31

dt = 67°C

Making c the subject of formula, we have;

`c = \frac {Q}{mdt}`

Substituting into the equation, we have;

`c = \frac {685}{7.9*67}`

`c = \frac {685}{529.3}`

Specific heat capacity, = 1.2942 J/g°C