Question

$800calories$ of heat is required to raise the temperature of $0.080kg$of a liquid from $10°C$to $100°C$. Find its specific heat capacity

(a) in calories (b) in joules.

Answer 1

Step 1: Given data

Mass$\left(m\right)=0.080kg$

Heat absorbed$=800calories$

The temperature change from $10°C$ to $100°C$

Specific heat capacity ($c$)=?

---

Step 2: (a)Specific heat capacity in calories

Temperature change$$\begin{gathered} =(100-10)°C \\ =90°C \end{gathered}$$

Heat absorbed$=mc{\theta }_R$

$$\begin{gathered} \Rightarrow 800calories=0.080kg\times c\times 90°C \\ \Rightarrow c=\frac{800}{0.080\times 90} \\ \Rightarrow c=111.1calk{g}^{-1}°C^{-1} \end{gathered}$$

---

Step 3:(b) Specific heat capacity in joules

To change into joule we will multiply the result in calories by$4.2$.

Specific heat capacity in joules$$\begin{gathered} =111.1\times 4.2 \\ =466.67Jk{g}^{-1}°C^{-1} \end{gathered}$$

---

The specific heat capacity in calories is $111.1calk{g}^{-1}°C^{-1}$and in joules is $466.67Jk{g}^{-1}°C^{-1}$.