Question

$100g$ of water is heated from ${30}^{o}C$ to ${50}^{o}C$ ignoring the slight expansion of the water, the change in its internal energy (Specific heat of water is $\left(4184J/Kg/K\right)$

• A. $4.2kJ$
• B. $8.4kJ$
• C. $84kJ$
• D. $2.1kJ$

Answer 1

The correct option is B $8.4kJ$

2) Ans By using heat energy change

$\Delta Q=MC\Delta T$

Where, m is the mass in kg

C is the specific heat

& $\Delta T$ is change in Temperature

$m=100g=100\times {10}^{-3}kg$

$C=4184J/kg/k$

$\Delta T=(50-30)={20}^{\circ }C$

So,

$\Delta Q=100\times {10}^{-3}\times 4184\times 20$

$\Delta Q=8400J$

$\Delta Q=8.4KJ$

By. law of thermodynamic

$\Delta U=\Delta Q+\Delta W$

$\Delta U=\Delta Q$ $\Delta W=$ change in work done $=0$

$B)\Delta U=8.4kJ$