Internal Energy and Energy Transfers

Solution

1. 1

   Heating supplies energy.

   $\text{Energy enters the water and increases its internal energy.}$

2. 2

   Effect on particles.

   $\text{Particles move faster — average kinetic energy increases.}$

3. 3

   Temperature.

   $\text{Temperature is a measure of average particle kinetic energy, so it rises.}$

Solution

1. 1

   Energy goes to potential.

   $\text{All input energy is used to break the bonds between water molecules.}$

2. 2

   Kinetic stays same.

   $\text{The average particle kinetic energy does NOT rise during the state change.}$

3. 3

   Conclude.

   $\text{Temperature depends on average kinetic energy, so it stays at 100 °C until all the liquid has become gas.}$

Solution

1. 1

   Δθ.

   $\Delta\theta = 100 - 22 = 78\,\text{°C}$

2. 2

   Apply formula.

   $\Delta E = 1.5 \times 4200 \times 78 = 491{,}400\,\text{J}$

Solution

1. 1

   Convert mass.

   $m = 0.200\,\text{kg}$

2. 2

   Δθ.

   $\Delta\theta = 80 - 25 = 55\,\text{°C}$

3. 3

   Apply formula.

   $\Delta E = 0.200 \times 4200 \times 55 = 46{,}200\,\text{J}$

Solution

1. 1

   Convert mass.

   $m = 0.500\,\text{kg}$

2. 2

   Δθ.

   $\Delta\theta = 84 - 22 = 62\,\text{°C}$

3. 3

   Rearrange.

   $c = \Delta E / (m \Delta\theta) = 15{,}500 / (0.500 \times 62) = 15{,}500 / 31 = 500\,\text{J/kg°C}$

Solution

1. 1

   Apply E = mL_f.

   $E = 0.25 \times 334{,}000 = 83{,}500\,\text{J}$

Solution

1. 1

   Convert mass.

   $m = 0.050\,\text{kg}$

2. 2

   Apply E = mL_v.

   $E = 0.050 \times 2{,}260{,}000 = 113{,}000\,\text{J}$

Solution

1. 1

   Convert mass.

   $m = 0.100\,\text{kg}$

2. 2

   Melt the ice.

   $E_1 = 0.100 \times 334{,}000 = 33{,}400\,\text{J}$

3. 3

   Heat water 0 to 100.

   $E_2 = 0.100 \times 4200 \times 100 = 42{,}000\,\text{J}$

4. 4

   Boil it.

   $E_3 = 0.100 \times 2{,}260{,}000 = 226{,}000\,\text{J}$

5. 5

   Total.

   $E = 33{,}400 + 42{,}000 + 226{,}000 = 301{,}400\,\text{J}$