Q62PE - College Physics (Urone)

Question

Integrated Concepts

How much time is needed for a surgical cauterize to raise the temperature of \(1.00{\rm{ }}\;g\) of tissue from \(37.0^\circ C\) to \(100^\circ C\) and then boil away \(0.500{\rm{ }}g\) of water, if it puts out \(2.00\;{\rm{ }}mA\) at \(15.0{\rm{ }}kV\)? Ignore heat transfer to the surroundings.

 

Step-by-Step Solution

Answer

The required time is \({\rm{45}}{\rm{.0\;s}}\).

Step 1: Definition of Power:

The power of a process is the amount of some type of energy converted into a different type divided by the time interval \({\rm{\Delta t}}\) in which the process occurred:

\({\rm{P}} = \frac{{{\rm{\Delta E}}}}{{{\rm{\Delta t}}}}\)                                                                                                         ….. (1)

 

The SI unit of power is the watt.

 

As you know, one watt is equal to one joule per second. Therefore, 

 1 W = 1 J/s 

The power of any electric circuit is defined by the formula

\({\rm{P}} = {\rm{I\Delta V}}\)                                                                                                          ….. (2)

Here, \({\rm{I}}\) is the current and \({\rm{I\Delta V}}\) is the potential difference.

Step 2: Definition of the thermal heat and Energy to boil or condense:

The thermal heat is defined by,

\({\rm{Q}} = {\rm{mc\Delta T}}\)                                                                                                     ….. (3)

Here, \({\rm{m}}\) is the mass of the substance, \({\rm{c}}\) is the specific heat of the substance and \({\rm{\Delta T}}\) is the change of temperature.

 

The energy required to vaporize or released of the gas condenses,

\({\rm{Q}} = {\rm{ \pm  m}}{{\rm{L}}_{\rm{v}}}\)                                                                                                      ….. (4)

Here, \({\rm{m}}\) is the mass of the liquid or gas, \({{\rm{L}}_{\rm{v}}}\) is the latent heat.

Step 3: The given data:

The mass of the tissue is:

\(\begin{align}{m_{tissue}} &= \left( {1.00{\rm{ }}g} \right)\left( {\frac{{1\;{\rm{ }}kg}}{{1000{\rm{ }}g}}} \right)\\ &= 1.00 \times {10^{ - 3}}{\rm{ }}kg\end{align}\).

The change in temperature of the tissue is: 

\(\begin{align}\Delta T &= 100^\circ C - 37.0^\circ C\\ &= 63.0^\circ C\end{align}\)

The mass of the boiled water is:

\(\begin{align}{m_{water}} &= \left( {0.500{\rm{ }}g} \right)\left( {\frac{{1\;{\rm{ }}kg}}{{1000{\rm{ }}g}}} \right)\\ &= 0.500 \times {10^{ - 3}}\;{\rm{ }}kg\end{align}\).

The current put out by the cauterize is:

\(\begin{align}I &= \left( {2.00\;mA} \right)\left( {\frac{{1\;A}}{{1000\;mA}}} \right)\\ &= 2.00 \times {10^{ - 3}}{\rm{ }}A\end{align}\).

- The voltage of the cauterize is:

\(\begin{align}\Delta V &= \left( {15.0{\rm{ }}kV} \right)\left( {\frac{{1000{\rm{ }}V}}{{1{\rm{ }}kV}}} \right)\\ &= 15.0 \times {10^3}{\rm{ }}V\end{align}\)

The specific heat of human body at \({\rm{37}}{\rm{.0}}^\circ {\rm{C}}\), 

\({L_v} = 2256 \times {10^3}{\rm{ }}{J \)

 

Step 4: Calculation of the power:

The output power of the cauterize from Equation (1) is,

\(P = I\Delta V\)

 

Substituting the known values in the above equation.

\(\begin{align}P &= \left( {2.00 \times {{10}^{ - 3}}{\rm{ }}A} \right)\left( {15.0 \times {{10}^3}{\rm{ }}V} \right)\\ &= 30.0{\rm{ }}W\end{align}\)

 

The required power is \(30.0{\rm{ }}W\).

Step 5: Calculation of the thermal energy

To raise the temperature of \({{\rm{m}}_{{\rm{tissue }}}}\), the energy required, from Equation (3):

\({Q_1} = {m_{tissue{\rm{ }}}}c\Delta T\)

 

Substituting the known values in the above equation.

\(\begin{align}{Q_1} &= \left( {1.00 \times {{10}^{ - 3}}{\rm{ }}kg} \right)\left( {3500{\rm{ }}{J \mathord{\left/

 {\vphantom {J {kg^\circ C}}} \right.

\kern-\nulldelimiterspace} {kg^\circ C}}} \right)\left( {63.0^\circ C} \right)\\ &= 220.5{\rm{ }}J\end{align}\)

 

To boil \({{\rm{m}}_{{\rm{water }}}}\), the energy required , from Equation (4):

\({Q_2} = {m_{water{\rm{ }}}}{L_v}\)

 

Substituting the values

\(\begin{align}{Q_2} &= \left( {0.500 \times {{10}^{ - 3}}{\rm{ }}kg} \right)\left( {2256 \times {{10}^3}{\rm{ }}{J \mathord{\left/

 {\vphantom {J {kg}}} \right.

 \kern-\nulldelimiterspace} {kg}}} \right)\\ &= 1128{\rm{ }}J\end{align}\)

 

The require thermal energy is \({\rm{1128 J}}\).

Step 6: Calculation of the time required:

The energy supplied by the cauterize is,

\({\rm{E}} = {{\rm{Q}}_{\rm{1}}}{\rm{ + }}{{\rm{Q}}_{\rm{2}}}\)

 

Substitute \({\rm{P\Delta t}}\) for \({\rm{E}}\) in the above equation.

\(P\Delta t = {Q_1} + {Q_2}\)

\(\Delta t = \frac{{{Q_1} + {Q_2}}}{P}\)

 

Substitute numerical values:

\(\begin{align}\Delta t &= \frac{{220.5\;{\rm{ }}J + 1128{\rm{ }}\;J}}{{30.0{\rm{ }}{J \mathord{\left/

 {\vphantom {J s}} \right.

 \kern-\nulldelimiterspace} s}}}\\ &= 45.0{\rm{ }}s\end{align}\)

 

Therefore, the required time is \(45.0{\rm{ }}s\).