GingerAle

Explain why it takes longer to cook a large cheese pizza (400g) than it does to cook a small cheese pizza (150g). Justify your answer with information on the relationship between the mass and thermal energy needed to change the temperature of an object.

Here's another: There are two automobiles. One has a mass of 1500kg the other has a mass of 4000kg. your mission is to  drive each automobile for 50 kilometers at 50 kph on the same course. You are to test if the more massive automobile takes longer to arrive at the destination compared to the less massive automobile.  Describe the relationship between the mass and energy needed to change the speed of the automobiles.  

I know it will take more energy, but will it take more time? Both automobiles are equipped with magic mushroom detectors.  Will one or both automobiles detect the presence of magic mushrooms?   If the time varies by more than few seconds, I suspect both detectors will indicate a positive...

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The pizza chef and Melody are correct: pizzas of different masses (with the same uniform density) cook at the same rate when exposed to the same uniform temperature. An assumption that time also increases is not in evidence (by even by careful observation) and makes this a poorly constructed question for introductory Newtonian physics relating to energy transfer.

The question asks for analysis for a time parameter via the relationship between mass and thermal energy. The question assumes a close to linear proportion of cooking time that’s related to the mass of the pizza, but this is negligible when cooking pizzas in a conventional oven.

The surface area of a pizza is directly proportional to its mass. So doubling the mass of a pizza also doubles its surface area. So, a larger pizza will absorb more total energy to cook but will not require more time to do this. The increased surface area is exposed to a larger reserve of energy (the heat in the oven). The reserve is depleted faster, causing the temperature to drop, which is detected by the temperature sensor that turns on an energy source to replenish the reserve. So, more energy is used in the same amount of time. (There may be a very small increase in cook time –a few seconds for the larger pizza because of the temperature threshold of the senor and its related delay in signaling for more energy, and the time needed to restore this energy to optimum levels.) 

It’s notable that multiple units of any food of the same mass and shape will not significantly augment the cooking time in a conventional oven. This assumes the multiple units do not directly interfere with the source of radiant heat, such as placing a pizza on a rack directly below another pizza. 

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Cooking spheroid shaped foods: Comparing the cooking process of a spherical beef roast to a circular beef steak with the same mass.

 It’s notable that while in increase in mass will always require a proportional increase in energy, changing only the shape (while keeping the same mass) does not require an increase of energy per se. Id est, it does not require an intrinsic increase of energy to cook the food. It will, however, require an increase in time for the energy to reach its target. Because of this extended time, more energy is lost to the outer environment and the energy source will have to replenish this loss to maintain the cooking temperature.

Consider a flat circular beefsteak and a spheroid roast same mass of 680 grams placed in a hypothetical oven that has uniform energy density distribution and is perfectly sealed to prevent energy loss.  Initially, both are exposed to an equal amount of energy. The steak, with the larger surface area, will absorb this energy much faster than the roast and finish the cooking process earlier. The difference in time corresponds to the ratios of the surface area per unit volume.

So here is a change in cooking time that’s completely independent of mass, but very dependent on surface area. 

A simplified example:

A 1.5lb spherical beef roast will have a mass of 680 grams and occupy a volume of (680g/1.02g/cc =) 667cc. This spherical volume will have a radius of 5.43cm and a surface area of 371cm^2.

A 1.5lb circular beef steak has the same mass (680g) and volume (667cc). With a 1.9cm (¾”) thickness the radius will be 10.57 and a total surface area of 828cm^2

Comparison of energy transfer for the above two (2) examples of beef:

Examination of the cooked beef will confirm the outer volumes (shells) of the beef will have absorbed and processed larger amounts of energy than the inner volumes –this energy has to pass through the outer shells to reach the inner shells. The differences in the absorbed energy is apparent in both the steak and roast, but is most pronounced in the roast

Generally, the cooking time for a spherical roast compared to a flat steak of the same mass is proportional to the surface area. The ratio of the surface areas is 2.23:1. Comparing recipes for various beef roasts suggest 18 to 20 minutes per pound with progressively elevated temperatures for rare, medium and well done roast. For the above spherical beef roast baked (in a preheated oven) at 350F for 27 to 30 minutes will produce a rare to medium rare center.  Cooking the circular beef stake at 350 for 12 to 13.5 minutes will produce a medium rare to medium stake. These times are close to the 2.23:1 ratio.

Below are three energy (heat) transfer formulas. I’m not going to elaborate on these beyond pointing out that all three have surface Area as part of the formula.   For any who are interested, Wikipedia has basic explanations for the dimensions and overviews for their uses.
 

*Newton’s law of cooling:

\(\dot Q = hA(T(t)) - T_\text{env}) = hA\,\Delta T(t) \)

Where ...

\(\dot {Q} \) is the rate of heat transfer out of the body.

\(h\) is the heat transfer coefficient (assumed independent of T and averaged over the surface) (SI unit: W/m2⋅K),

\(A\) is the heat transfer surface area (SI unit: m2),

\(T\) is the temperature of the object's surface (SI unit: K)

\({\displaystyle T_{\text{env}}} \)is the temperature of the environment (SI unit: K).

\({\displaystyle \Delta T(t)=T(t)-T_{\text{env}}}\)is the time-dependent temperature difference between environment and object (SI unit: K).

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*Convective heat transfer:  \(\dot {Q} = hA(T-T_{f})\)

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*Radiant heat transfer:  \( \displaystyle {\dot {Q_r}} = -hrA(T1 - T2)\)

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For comparison, it’s worth noting that an increase in the mass of food will affect cooking time requirements in a microwave oven, but the increase in time is not linear to the increase in food mass. The reason for this increase in time is because a typical domestic microwave operating at 1100 watts of input power (at 80% magnetron efficiency) outputs about 3000 BTUH of power into the chamber. This is trivial compared to the 16000 to 18000 BTUH (4.8 to 5.2KW) produced by a gas or electric oven, where more than 99% of this energy is wasted.   

The energy inside a microwave chamber is wasted too, but it’s a much smaller percentage than for a conventional oven. (The microwave energy bounces around the chamber until it’s absorbed by the food or the chamber itself. More food means less loss in the chamber, but only to a point.)

A medium sized potato requires about 75 watt*hours (±12%) to fully cook. An 1100 watt microwave will transmit this energy to the potato in ~5 minutes, using about 95 watt*hours of energy. A conventional oven requires ~50 minutes to transfer 75 watt*hours to an identical potato, and uses 5000 watt*hours to do this. This incredible waste of energy costs about 80 cents, of which about 1.5 cents goes into the potato (based on U.S mean of 20 cents per kilowatt*hour of electric energy) plus taxes and surcharges. For natural gas, the cost is about 30 cents, of which 0.6 cents goes into the potato. (based on U.S mean of 174 cents per Therm). However the oven can be stuffed with as many potatoes that can fit on a rack with only a trivial increase in energy requirements. 

In a microwave oven, if a medium sized potato requires five (5) minutes to cook, adding an another (identical) potato will require an additional two (2) minutes for a total of about seven (7) minutes. This implies in increase in efficiency of about 30%, based on cooking time per unit mass.  

Adding a third potato also requires an additional two (2) minutes, and this increases efficiency of  about 55%. At some point (about five or more potatoes), the cook time will start to significantly increase. With eight potatoes, the distributed energy will become too low to heat the individual potatoes enough to cook them. The potatoes lose heat faster than it can be replaced.

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Now that I’ve finished this tortuous treatise, I’m considering a tempting treat.  A decision on whether to cook up a pizza, a round roast, a round steak, or a potato.  I think the steak with a potato sounds good. I’m going to broil the steak though. (Broil is an American English word that means to grill under a flame or element.) It’s a wonderful way to cook a steak –not as good as a charcoal barbeque grill though.  

Next week, I’m going to throw a few shrimps and prawns on the Barbie. --they are actually two different suborders of decapods on the crustacean family tree. I'll casually check if they cook at different rates.  

 I suspect I’ll be remembering this post. 

GA 

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Solution:

\(\mathrm G_1 = \text {cost of gown 1}\\ \mathrm G_2 = \text {cost of gown 2 }\\ \text {Arbitrary sale price } $100 \\ ---\\ \mathrm G_1 + 0.25G_1 = 100 \\ 4(\mathrm G_1) + \mathrm G_1 = 400 \\ \mathrm 5G_1 = 400 \\ \mathrm G_1 = 80 \\ ---- \\ \)

\(G_2 + (-.025G_2) = 100 \\ 4(G_2) = - G_2 = 400 \\ 3G_2 = 400 \\ G_2 = 133.33 \\ \text {Total cost for gowns } = 213.33\\ \text {Revenue from sale }= 200.00 \\ \text {P/(L) } = (13.33)\\ \text {P/(L)%} = (1-(200/213.33)) \rightarrow (6.25\%) \text { Loss}\\ \)

Solution (without the use of an arbitrary sale price):

\(\mathrm G_1 + 0.25\mathrm G_1 = \mathrm G_2 - 0.25\mathrm G_2\\ 4\mathrm G_1 + \mathrm G_1 = 4\mathrm G_2 - \mathrm G_2 \\ 5\mathrm G_1 = 3\mathrm G_2 \\ \mathrm G_1 = 3/5\mathrm G_2 \\ \mathrm G_1 = 0.60\mathrm G_2 \\ --- \\ \mathrm G_1 \text { must equal 60% of the cost of } \mathrm G_2\\ \text {Or } \\ \mathrm G_2 \text { must equal 167% of the cost of } \mathrm G_1\\ \)

GA

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The selling price was not 1.25x or 0.75x, it was the ending cost because if they made "25%" off of one of the gowns, the profit would've been 25$, the starting price wouldn't have been 1.25x... That's a logic flaw...

Proyaop, there is no logical error in the question or Mr. BB’s solution.

Note the question:

Martha's millinery sold two evening gowns at the same price. Based on their cost, the store made 25% on one of the gowns and lost 25% on the other. ...

The gowns sell at the same price. The profit and loss is based on the (store’s) costs for the gowns. 

So, Mr. BB’s nonstandard (slop) presentation depicts the correct solution.

GA

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What happened to the -1/9 ?

Where did the 1/8 come from ?

The graph and related text is a separate question. It is not an attempt to answer the (first) question.  There are two (2) questions in the above post.  

Twiistedwhyami, one reason for your twisted thought process is the lack of understanding for the necessary prerequisites for the math questions you presented. You should request a return to the prior, lower level math class. The teacher gave you passing marks to get rid of you. You didn’t really pass.

You should also request a return to your prior English class: your statement, “no typo, it came with the problem” is fragmented and lacks proper punctuation.

If you do not know how to communicate how do you expect to learn anything?   

GA

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