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Internal Energy

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Question

Find the change in the internal energy (in $J$ ) of $2 k g$ of water as it is heated from $0^{o} C t o 4^{o} C$ . The specific heat capacity of water is $4200 J / k g - K$ and its densities at $0^{o} C$ and $4^{o} C$ are $999.9 k g / m^{3} a n d 1000 k g / m^{3}$ respectively. Atmospheric pressure = $10^{5} P a$ .

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Solution

form the first law of thermodynamics
$d Q = d U + d W$
work done is not completely zero but it's magnitude is very small so it can be neglected,or we can find it applying isobaric process.
$Q = m S Δ T$
$Q_{1} = 2 \times 4200 \times 273 = 2293200 J$
$Q_{2} = 2 \times 4200 \times 277 = 2326800 J$
now $Δ U = Q_{2} - Q_{1} = 33600 J$

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Similar questions

10 g

0^{\circ} t o 100^{\circ}

100 k P a

x \times 10^{2} J

x

= 0.6 k g / m^{3} .

= 4200 J / k g -^{\circ} C

= 2.5 \times 10^{6} J / k g .

=

=

2 K g

0^{0} C

20^{0} C

3, 34, 000 J / K g

4200 J {K g}^{- 1} K^{- 1}

20 g

0^{\circ} C

100^{\circ} C

100 kPa

= 0.50 {kg/m}^{3}

= 4200 {J kg}^{- 1}^{\circ} C^{- 1}

= 2.3 \times 10^{6} J/kg

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