Question

7 L of water at 323 K is taken into 2 beakers shown below as 5 L and 2 L. If the water in each beaker is given 1000 calorie heat each, find the change in temperature in each beaker. (Ignore the heat lost to surroundings)

• A. Beaker A - 5 K ; Beaker B - 2.5 K
• B. Beaker A - 2 K ; Beaker B - 5 K
• C. Beaker A - 2.5 K ; Beaker B - 5 K
• D. Beaker A - 5 K ; Beaker B - 2 K

Answer 1

The correct option is B Beaker A - 2 K ; Beaker B - 5 K

We know, specific heat capacity of water = $4200J/KgK$
Density of water = $\rho =1kg/L$
Also $1cal=4.2J$
hence, $10000cal=42000J$
Given
Volume of water in beaker A $=\text{V}=5\text{L}$
Mass of water in beaker A = (V$\times \rho$)
$=5kg$
Volume of water in beaker B = $\text{V}=2\text{L}$
Mass of water in beaker B = $(\text{V}\times \rho )$
$=2kg$
Using the equation $\text{Q}=\text{m c}\theta$ , where Q = Quantity of heat required to rise the temperature.
C = Specific heat capacity
$\theta$ = Change in temperature

Beaker A
Mass = $5kg$
Change in temperature ($\theta$) = $\frac{\text{Q}}{\text{m c}}=\frac{42000}{5\times 4200}=2K$

Beaker B
Mass = 2 kg
Change in temperature ($\theta$) = $\frac{Q}{mc}=\frac{42000}{2\times 4200}=5K$

Hence the change in temperature in the beaker A is $2K$ and in the beaker B is $5K$.