Question

Find the increase in mass when 1 kg of water is heated from 0°C to 100°C. Specific heat capacity of water = 4200 J kg−1 K−1.

Answer 1

THREE STEPS SOLUTIONS

Explanation:

Given,

Mass of water, m = 1 kg

Specific heat capacity of water = 4200 $j kg^{-1} k^{-1}$

Change in temperature , $\Delta\theta$ = 100°C

Required heat energy, Q = ms\Delta\theta

Q = $1\times 4200\times 100$

Q = 420000 J

The energy is converted into mass,

so, increase in mass of water on heating,

$\Delta m = \frac{Q}{C^{2}}$

$\Delta m =\frac{420000}{({3\times{10}^{8}})^{2}}$

$\Delta m = \frac{42\times 10^{4}}{9\times 10^{16}}$

$\Delta m = 4.66 \times 10^{-12} kg$

Answer 2

Increase in Mass is $4.7 \times 10^{-12}$ kg

Explanation:

Given,

$\Delta T$ = 100 - 0 = 100°C

$m_0$ = 1 kg

S = $4200\ jkg^-^1. K$

• Thus amount of heat required to raise the temperature,

Q = $mS (\Delta T)$

= $1 \times 4200 \times 100$

= $4.2 \times 10^5$ J

• Using Einstein's mass-energy equivalence,  

$E = mc^2$

where m is the increase in mass

So, $4.2 \times 10^5 = m \times (3 \times 10^8)^2$

Hence, $m = 4.7 \times 10^{-12}\ kg$