Question

Suppose you warm up 520 grams of water (about half a liter, or about a pint) on a stove, and while this is happening, you also stir the water with a beater, doing 5multiply.gif104 J of work on the water. After the large-scale motion of the water has dissipated away, the temperature of the water is observed to have risen from 21°C to 84°C.
What was the change in the thermal energy of the water?

deltacapEthermal =?

Taking the water as the system, how much energy transfer due to a temperature difference (microscopic work) Q was there across the system boundary?
Q = ?

Taking the water as the system, what was the energy change of the surroundings?
deltacapEsurroundings=?

Answer 1

Answer:

$\Delta Q=137067.84\ J$ is the change in the thermal energy of water this is also the amount of energy crossing the system boundary due to temperature difference.

$\delta E=187067.84\ J$

Explanation:

Given:

• mass of water heated, $m=520\ g=0.52\ kg$

• work done on the water by stirring, $W=5\times 10^4\ J$

• initial temperature of water, $T_i=21^\circ{}$

• final temperature of water, $T_f=84^{\circ}$

Now the change in thermal energy of the water is depicted by the change in the temperature of the water.:

for water we've specific heat capacity, $c=4184\ J.kg^{-1}.^{\circ}C^{-1}$

so,

$\Delta Q=m.c.(T_f-T_i)$

$\Delta Q=0.52\times 4184\times(84-21)$

$\Delta Q=137067.84\ J$ is the change in the thermal energy of water this is also the amount of energy crossing the system boundary due to temperature difference.

The energy change in the surrounding will be equal to the energy change in the system.

so,

$\delta E=\Delta Q+W$

$\delta E=137067.84+50000$

$\delta E=187067.84\ J$