Question

Shoveling snow can be extremely taxing because the arms have such a low efficiency in this activity. Suppose a person shoveling a footpath metabolizes food at the rate of\(800{\rm{ W}}\).

(a) What is her useful power output?

(b) How long will it take her to lift\(3000{\rm{ kg}}\)of snow\(1.20{\rm{ m}}\)? (This could be the amount of heavy snow on\(20{\rm{ m}}\)of footpath.)

(c) How much waste heat transfer in kilojoules will she generate in the process?

Short answer

(a) The useful power output of a person shoveling snow is \(24{\rm{ W}}\).

(b) Time taken by her to lift snow is \(24.5{\rm{ s}}\).

(c) The waste heat transfer generated by her process is \(34.2{\rm{ kJ}}\).

Step-by-step solution

Step 1: Definition of Concepts 

Efficiency: Efficiency is defined as the ratio of useful output by the total input.

Mathematically,

\(\eta \% = \frac{{{P_{{\rm{ouput}}}}}}{{{P_{input}}}} \times 100\% \)

Here, \({P_{output}}\) is the useful output power and \({P_{input}}\) is the total input power.

Find the useful power output of a person shoveling snow

(a)

The useful power output of the person shoveling snow can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the useful output power.

\({P_{output}} = \frac{{{P_{input}} \times \eta \% }}{{100\% }}\)

Here, \({P_{output}}\) is the useful output power, \({P_{input}}\) is the total input power \(\left( {{P_{input}} = 800{\rm{ W}}} \right)\), and \(\eta \) is the efficiency of the person shoveling snow \(\left( {\eta = 3\% } \right)\).

Putting all known values,

\(\begin{array}{c}{P_{output}} = \frac{{\left( {800{\rm{ W}}} \right) \times \left( {3\% } \right)}}{{100\% }}\\ = 24{\rm{ W}}\end{array}\)

Therefore, the useful power output of a person shoveling snow is \(24{\rm{ W}}\).

Calculate the time taken to lift snow

(b)

The work done in lifting snow is,

\(W = mgh\)

Here, \(m\) is the mass of snow \(\left( {m = 3000{\rm{ kg}}} \right)\), \(g\) is the acceleration due to gravity \(\left( {g = 9.81{\rm{ m}}/{{\rm{s}}^2}} \right)\), and \(h\) is height \(\left( {h = 1.20{\rm{ m}}} \right)\).

Putting all known values,

\(\begin{array}{c}W = \left( {3000{\rm{ kg}}} \right) \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {1.20{\rm{ m}}} \right)\\ = 35280{\rm{ J}}\end{array}\)

The useful power output is used in doing work. The useful power output is,

\({P_{output}} = \frac{W}{t}\)

Here, W is the work done in lifting snow, and t is the time taken to lift the snow.

The expression for the time is,

\(t = \frac{W}{{{P_{output}}}}\)

Putting all known values,

\(\begin{array}{c}t = \frac{{35280{\rm{ J}}}}{{24{\rm{ W}}}}\\ = 1470{\rm{ sec}} \times \left( {\frac{{1{\rm{ min}}}}{{60{\rm{ sec}}}}} \right)\\ = 24.5{\rm{ min}}\end{array}\)

Therefore, the required time taken by her to lift snow is \(24.5{\rm{ s}}\).

Calculate the waste heat transfer generated by her process

(c)

Since the efficiency of shoveling snow is \(3\% \), which means \(97\% \) of the total work is wasted.

Therefore, waste heat transfer is,

\({W_{waste}} = W \times 97\% \)

Putting all known values,

\(\begin{array}{c}{W_{waste}} = \left( {35280{\rm{ J}}} \right) \times \left( {\frac{{97}}{{100}}} \right)\\ = 34221.6{\rm{ J}} \times \left( {\frac{{1{\rm{ kJ}}}}{{1000{\rm{ J}}}}} \right)\\ = 34.22{\rm{ kJ}}\end{array}\)

Therefore, the required waste heat transfer generated by her process is \(34.2{\rm{ kJ}}\).