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

Answer:

$c=2.0769\ \frac{J}{g\ \textdegree C}$

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:

$q=mc\bigtriangleup T\\\\c=\frac{q}{m\bigtriangleup T}\\\\\bigtriangleup T=(57-32)\textdegree C=25\textdegree C\\\\\therefore c=\frac{6.75\times 10^4J}{1.3\times 1000\ g\times 25\textdegree C}\\\\=2.0769 \ \frac{J}{g\ \textdegree C}$

Hence, the specific heat capacity of the ice is $2.0769 \frac{J}{g\ \textdegree C}$