Question

Calculate the specific heat capacity of a piece of ice if 1.30 kg of the wood absorbs 6.75×104 joules of heat, and its temperature changes from 32 ºC to 57 ºC.

Answer 1

`c=2.0769\ (J)/(g\ \textdegree C)`

Step-by-step explanation:

-Specific heat capacity is given by the formula:

`q=mc\bigtriangleup T`

Where:

`q` is the heat gained or loosed by the substance

`m` is the mass of the substance

`c` is the specific heat of the substance

`\bigtriangleup T` is the change in temperature

#We make c the subject of the formula and substitute to solve for it:

`q=mc\bigtriangleup T\\\\c=(q)/(m\bigtriangleup T)\\\\\bigtriangleup T=(57-32)\textdegree C=25\textdegree C\\\\\therefore c=(6.75* 10^4J)/(1.3* 1000\ g* 25\textdegree C)\\\\=2.0769 \ (J)/(g\ \textdegree C)`

Hence, the specific heat capacity of the ice is
`2.0769 (J)/(g\ \textdegree C)`