Question

Heat energy is given to $100g$ of water, such that its temperature rises by $10K$. When the same heat energy is given to the liquid L of mass $50g$ its temperature rises by $50K$. Calculate (a) Heat energy given to water (b) Specific heat capacity of liquid L {take specific heat capacity of water =$4200JK{g}^{-1}{K}^{-1}$]

Answer 1

Step 1

Given:

1. Mass of water 100g, temperature rise by 10 K
2. Mass of Liquid L 50g, temperature rise by 50 K

Step 2

Find:

(a) Heat energy is given to water

(b) Specific heat capacity of liquid L

Formula used:

Heat energy = $mc{\theta }_R$

Step 3

Heat energy is given to water

$$\begin{gathered} =mc{\theta }_R \\ =100\times 4.2\times 10[c\to 4.2J{g}^{-1}{K}^{-1}] \\ =4200J \end{gathered}$$

Step 4

Heat energy given to liquid L is $4200J$

We have,

Heat energy is given to liquid L =$mc{\theta }_R$

$$\begin{gathered} 4200=mc{\theta }_R \\ 4200=50\times c\times 50 \\ c=\frac{4200}{2500} \\ c=1.68J{g}^{-1}{K}^{-1} \\ c=1680JK{g}^{-1}{K}^{-1} \end{gathered}$$

Hence,

(a) Heat energy is given to water $4200J$

(b) Specific heat capacity of liquid L $1680JK{g}^{-1}{K}^{-1}$