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If I accidentally dropped my cooked steak off the deck into the snow, how much snow could it melt? Recall that for H,0, L, = 3.33 × 10° L, = 22.6 x 10° , and c = 4186 (not all values needed) b If instead I accidentally dropped my cooked steak in a bucket with 0.15 kg of snow (which instantly became sealed to the environment), what temperature would the steak reach?

If I accidentally dropped my cooked steak off the deck into the snow, how much snow could it melt? Recall that for H,0, L, = 3.33 × 10° L, = 22.6 x 10° , and c = 4186 (not all values needed) b If instead I accidentally dropped my cooked steak in a bucket with 0.15 kg of snow (which instantly became sealed to the environment), what temperature would the steak reach?

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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need help with parts a & b

#7. Suppose I am outside grilling on a 0 °C day. I grill a 0.64 kg steak for 7.0 min. Suppose my grill can transfer heat to my steak at 330 !. How do I like my steaks cooked? You can assume that although starting at 0 °C and composed mostly of water, there is no phase change for the steak. Note that the specific heat capacity for steak is 3490 e. Also note that a rare steak is cooked to above 130 °F, a meadium-rare steak is cooked to above 140 °F, a medium steak is cooked to above 150 °F, a medium-well steak is cooked to above 160 °F, and a well-done steak is cooked to above 165 °F. we will solve this by basic calorimetry , once we get the final temperature we can guess the type of stake e = (T; – r;) Q = heat transferred m c = 330 * 60 * 7 m = mass of steak = 0. 64 c = 3490 T = final temp ? = initial temp = 0 Ti + T; Ts me 330 *60*7 + 0 = T; 62. 05229 celcius = 143. 694122 fahrenheit 0.64 * 3490 so it is medium rare a If I accidentally dropped my cooked steak off the deck into the snow, how much snow could it melt? Recall that for H;O, L1 = 3.33 x 10*¿L= 22.6 × 10° , and c = 4186 (not all values needed) b If instead I accidentally dropped my cooked steak in a bucket with 0.15 kg of snow (which instantly became sealed to the environment), what temperature would the steak reach?
Transcribed Image Text:#7. Suppose I am outside grilling on a 0 °C day. I grill a 0.64 kg steak for 7.0 min. Suppose my grill can transfer heat to my steak at 330 !. How do I like my steaks cooked? You can assume that although starting at 0 °C and composed mostly of water, there is no phase change for the steak. Note that the specific heat capacity for steak is 3490 e. Also note that a rare steak is cooked to above 130 °F, a meadium-rare steak is cooked to above 140 °F, a medium steak is cooked to above 150 °F, a medium-well steak is cooked to above 160 °F, and a well-done steak is cooked to above 165 °F. we will solve this by basic calorimetry , once we get the final temperature we can guess the type of stake e = (T; – r;) Q = heat transferred m c = 330 * 60 * 7 m = mass of steak = 0. 64 c = 3490 T = final temp ? = initial temp = 0 Ti + T; Ts me 330 *60*7 + 0 = T; 62. 05229 celcius = 143. 694122 fahrenheit 0.64 * 3490 so it is medium rare a If I accidentally dropped my cooked steak off the deck into the snow, how much snow could it melt? Recall that for H;O, L1 = 3.33 x 10*¿L= 22.6 × 10° , and c = 4186 (not all values needed) b If instead I accidentally dropped my cooked steak in a bucket with 0.15 kg of snow (which instantly became sealed to the environment), what temperature would the steak reach?
Expert Solution
Step 1

maximum snow will melt when the steak will reach 0 celcius

and the heat released will be used as latent heat.

Q = m c (Tf- Ti)  = mass of snow *  latent heat 0.64*3490* (62.05229-0)=330*60*7 =mass of snow * 3.33 * 100000mass of snow = 330*60*73.33 * 100000 = 0.416216216 kg

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