Chapter 22.2, Problem 29PP

a.

To determine

To Find:The current through the stove element.

Answer to Problem 29PP

  $I=20\text{A}$

Explanation of Solution

Given:

Resistance, $\text{R=11}\Omega$

Voltage, $\text{V=220 V}$

Formula Used:

Ohm’s law

  $V=IR$

V is the voltage

I is the current

R is the resistance

Calculation:

Using Ohm’s law

  $I=\frac{\text{V}}{\text{R}}$

Putting the values

  $I=\frac{\text{220 V}}{\text{11}\Omega }$

  $I=20\text{A}$

Conclusion:

Current through the stove element, $I=20\text{A}$.

b.

To determine

The transformed thermal energy in $30\text{ s}$ .

Answer to Problem 29PP

  $\text{E=1}\text{.3}\times {10}^5\text{J}$

Explanation of Solution

Given:

Voltage, $V=\text{220 V}$

Resistance, $R=\text{11}\Omega$

Current, $I=20\text{A}$

Time, $t=30\text{ s}$

Formula Used:

  $E=I^{2}Rt$

E is the energy

I is the current

R is the resistance

t is the time.

Calculations:

  $E=I^{2}Rt$

Putting the values

  $E={(20\text{A)}}^2\times \text{11}\Omega \times 30\text{ s}$

  $E=\text{1}\text{.3}\times {10}^5\text{J}$

Conclusion:

Thermal energy transformed is $E=\text{1}\text{.3}\times {10}^5\text{J}$

c.

To determine

The rise in temperature of water.

Answer to Problem 29PP

  $\Delta T=17°\text{C}$

Explanation of Solution

Given:

Mass of water, $m=\text{1}\text{.2}\text{kg}$

Specific heat capacity of water, $c=\text{4180 J/kg}\cdot °\text{C}$

  $E=\text{1}\text{.3}\times {10}^5\text{J}$

Formula Used:

  $Q=mc\Delta T$

Q is the heat supplied

m is the mass of water

c is the specific heat capacity of water.

Calculations:

Given

  $Q=\text{0}\text{.65}\times E=\text{0}\text{.65}\times \text{1}\text{.3}\times {10}^5\text{J}$

  $\Delta T=\frac{mc}{Q}$

Putting the values

  $\Delta T=\frac{\text{1}\text{.2kg}\times \text{4180 J/kg}\cdot °\text{C}}{\text{0}\text{.65}\times \text{1}\text{.3}\times {10}^5\text{J}}$

  $\Delta T=17°\text{C}$

Conclusion:

Rise in temperature, $\Delta T=17°\text{C}$