Question

A 33.05 g sample of a substance is initially at 28.5 °C. After absorbing 2589 J of heat, the temperature of the substance is 107.9 'C.

What is the specific heat (c) of the substance?

Answer 1

`0.985\ \text{J/g}^(\circ)\text{C}`

Step-by-step explanation:

m = Mass of sample = 33.05 g

`\Delta T` = Change in temperature =
`107.9-28.5=79.5^(\circ)\text{C}`

Q = Heat absorbed = 2589 J

c = Specific heat of substance

Heat is given by

`Q=mc\Delta T\\\Rightarrow c=(Q)/(m\Delta T)\\\Rightarrow c=(2589)/(33.05* 79.5)\\\Rightarrow c=0.985\ \text{J/g}^(\circ)\text{C}`

The specific heat of the substance is
`0.985\ \text{J/g}^(\circ)\text{C}`.