Question

Wheen 56j of heat are added to 11g of liquid, its temperature rises from 10.4 degrees celsius to 12 degrees celsius. What is the heat capacity of the liquid

Answer 1

$c = 3.2 \; \text{J} \cdot \text{kg}^{-1} \cdot \text{K}^{-1}$

• Energy change $Q = 56 \; \text{J}$
• Mass being heated $m = 11 \; \text{g}$
• Temperature change $\Delta T = 12 - 10.4 = 1.6\;^{\text{o}}\text{C}$ which is the same as $1.6 \;\text{K}$.

Heat capacity measures the energy required to raise the temperature of a unit mass of a substance by a unit degree. Therefore

$c = Q / (m \cdot \Delta T) = 56/(11 \times (12 - 10.4)) = 3.2 \; \text{J} \cdot \text{kg}^{-1} \cdot \text{K}^{-1}$

Answer 2

Answer : The heat capacity of liquid will be, $3.18J/g^oC$

Explanation :

Formula used :

$Q=m\times c\times \Delta T\\\\Q=m\times c\times (T_2-T_1)$

where,

Q = heat absorb = 56 J

m = mass of liquid = 11 g

c = specific heat of liquid = ?

$\Delta T$ = change in temperature

$T_1$ = initial temperature = $10.4^oC$

$T_2$ = final temperature = $12^oC$

Now put all the given value in the above formula, we get:

$56J=11g\times c\times (12-10.4)^oC$

$c=3.18J/g^oC$

Therefore, the heat capacity of liquid will be, $3.18J/g^oC$