Question

in spring, turkeys eat tender greens, shoots, tubers, leftover nuts and early insects. as the weather warms up, they eat more insects, including grasshoppers, walking - sticks, beetles, weevils, dragonflies, ants and larvae. they also consume spiders, harvestmen, ticks, millipedes, centipedes, snails and slugs. but even in summer, a majority of the diet (perhaps 90 percent) is vegetables. a wide variety of plant species are eaten, as well as a number of plant parts, including fruits, seeds, seedheads, tubers, roots, bulbs, stems, leaves, flowers and buds.
in fall, turkeys eat mast (beechnuts, acorns); fruits (dogwood, grape, cherry, gum, thornapple); and seeds (grasses and sedges, ash, corn, oats, weeds). during winter, they rely on seeds, nuts and fruits left over from autumn, and on green plants, crustaceans and insect larvae found in and around spring seeps where groundwater emerges along a hillside or in a flat. temperature of this water is above freezing, so the seeps remain open all winter, providing food for turkeys and other wildlife.
how many pounds of insects will a turkey poult eat in a day? the volume of food consumed by wild turkeys on a daily basis varies by season and abundance. many studies of wild turkey food habits expressed the volume in terms of cubic centimeters and grams rather than pounds. certainly the volume of insects that can be consumed by wild turkey poults changes rapidly as the poults grow. korschgen (1967) reported that domestic turkeys require about 13.6 grams of food per pound of body weight each day to sustain weight. wild turkeys are more active, but using this information a 10 pound wild turkey would require about 136 grams and a 20 pound gobbler would require 272 grams a day.
the typical male turkey has a mass of 17.5 lbs whereas the typical female turkey has a mass of 8.0 lbs. how many ounces of insects would 7 gobblers consume daily to sustain themselves? show your work using dimensional analysis.

Explanation:

Step1: Find the food consumption rate

From the text, wild turkeys (gobblers are male, so we use the 13.6 grams per pound of body weight per day rate? Wait, no, wait: "domestic turkeys require about 13.6 grams of food per pound of body weight each day to sustain weight. Wild turkeys are more active, but using this information a 10 pound wild turkey would require about 136 grams and a 20 pound gobbler would require 272 grams a day." Wait, let's check that. For a 10 - pound wild turkey: 10 pounds * 13.6 grams/pound = 136 grams. For a 20 - pound gobbler: 20 pounds * 13.6 grams/pound = 272 grams. So the rate is 13.6 grams of food per pound of body weight per day.

Step2: Calculate for one 17.5 - pound gobbler

First, find the daily food consumption for one gobbler. The mass of a male turkey (gobbler) is 17.5 lbs. Using the rate of 13.6 grams per pound of body weight per day. So the amount of food for one gobbler per day is \(17.5\space lbs\times13.6\space\frac{g}{lb}\). Let's calculate that: \(17.5\times13.6 = 238\space grams\) per day for one gobbler.

Step3: Calculate for 7 gobblers

Now, we need to find the amount for 7 gobblers. So multiply the amount for one gobbler by 7: \(238\space g\times7=1666\space grams\) per day for 7 gobblers.

Step4: Convert grams to ounces

We know that 1 ounce is approximately 28.35 grams. So to convert grams to ounces, we use the conversion factor \(\frac{1\space oz}{28.35\space g}\). So the number of ounces is \(1666\space g\times\frac{1\space oz}{28.35\space g}\). Let's calculate that: \(\frac{1666}{28.35}\approx58.76\space oz\). Wait, but let's check the rate again. Wait, the text says "domestic turkeys require about 13.6 grams of food per pound of body weight each day to sustain weight. Wild turkeys are more active, but using this information a 10 pound wild turkey would require about 136 grams and a 20 pound gobbler would require 272 grams a day." Wait, 10 pound * 13.6 g/pound = 136 g, 20 pound * 13.6 g/pound = 272 g. So the rate is indeed 13.6 g per pound per day. So for a 17.5 - pound gobbler: 17.5 * 13.6 = 238 g per day. For 7 gobblers: 7 * 238 = 1666 g. Now convert grams to ounces. Since 1 ounce = 28.3495 grams (more accurate value). So \(1666\div28.3495\approx58.76\) ounces. If we use 28.35 as the conversion factor, it's approximately 58.8 ounces. But let's do the calculation more precisely.

Wait, maybe we made a mistake in the rate. Wait, the text says "Wild turkeys are more active, but using this information a 10 pound wild turkey would require about 136 grams and a 20 pound gobbler would require 272 grams a day." So for 10 pounds, 136 grams: 136/10 = 13.6 grams per pound. For 20 pounds, 272 grams: 272/20 = 13.6 grams per pound. So the rate is 13.6 grams per pound per day. So that part is correct.

So let's re - do the steps with correct unit conversions:

Alternative Step 2: Wait, maybe the problem is about insects? Wait, the question is "How many ounces of insects would 7 gobblers consume daily to sustain themselves?" Wait, is the food all insects? The text says "the volume of insects that can be...", but the consumption rate given is for food (which may include insects? Wait, the problem says "How many ounces of insects...", but the rate given is for food. Maybe we assume that the food consumption rate is for insects? The text is a bit ambiguous, but we'll proceed with the given rate.

So step 1: Rate \(r = 13.6\space\frac{g}{lb\cdot day}\)

Step 2: Mass of one gobbler \(m = 17.5\space lb\)

Step 3: Daily consumption for one gobbler \(C_1=m\times r=17.5\space lb\times13.6\space\frac{g}{lb}=238\space g/day\)

Step 4: Daily consumption for 7 gobblers \(C_7 = 7\times C_1=7\times238 = 1666\space g/day\)

Step 5: Convert grams to ounces. We know that \(1\space oz = 28.3495\space g\), so \(C_7(oz)=\frac{1666\space g}{28.3495\space g/oz}\approx58.8\space oz\) (rounded to one decimal place) or approximately 59 oz.

Answer:

Approximately \(\boldsymbol{58.8}\) ounces (or about 59 ounces)