{"id":87147,"date":"2025-06-01T04:30:08","date_gmt":"2025-06-01T04:30:08","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=87147"},"modified":"2025-06-01T04:30:08","modified_gmt":"2025-06-01T04:30:08","slug":"a-cup-of-tea-is-placed-at-a-table-and-the-room-temperature-is-35c-t","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/a-cup-of-tea-is-placed-at-a-table-and-the-room-temperature-is-35c-t\/","title":{"rendered":"A cup of tea is placed at a table, and the room temperature is 35\u00b0C. T"},"content":{"rendered":"

A cup of tea is placed at a table, and the room temperature is 35\u00b0C. The tea cools from 90\u00b0C to 70\u00b0C in 5 minutes. Which one among the following is the correct time taken for the tea to cool from 70\u00b0C to 50\u00b0C ?<\/p>\n

[amp_mcq option1=”5 minutes” option2=”7 minutes” option3=”9 minutes” option4=”11 minutes” correct=”option3″]<\/p>\n

\n
\n
This question was previously asked in<\/div>\n
UPSC Geoscientist – 2024<\/div>\n<\/div>\n
Download PDF<\/a>Attempt Online<\/a><\/div>\n<\/div>\n
\nThe correct option is C.
\n<\/section>\n
\nAccording to Newton’s Law of Cooling, the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. That is, d\u03b8\/dt \u221d (\u03b8 – \u03b8\u2080), where \u03b8 is the object’s temperature and \u03b8\u2080 is the surrounding temperature. For small temperature ranges, the average rate of cooling in an interval can be approximated as proportional to the average temperature difference in that interval.
\n<\/section>\n
\nLet the surrounding temperature be \u03b8\u2080 = 35\u00b0C.
\nIn the first interval, the tea cools from 90\u00b0C to 70\u00b0C in t\u2081 = 5 minutes.
\nThe temperature change is \u0394\u03b8\u2081 = 90 – 70 = 20\u00b0C.
\nThe average temperature in this interval is \u03b8_avg\u2081 = (90 + 70) \/ 2 = 80\u00b0C.
\nThe average temperature difference is \u0394T\u2081 = \u03b8_avg\u2081 – \u03b8\u2080 = 80 – 35 = 45\u00b0C.
\nThe average rate of cooling in this interval is approximately Rate\u2081 \u2248 \u0394\u03b8\u2081 \/ t\u2081 = 20\u00b0C \/ 5 min = 4\u00b0C\/min.
\nBy Newton’s Law of Cooling, Rate\u2081 \u221d \u0394T\u2081, so 4 \u221d 45.<\/p>\n

In the second interval, the tea cools from 70\u00b0C to 50\u00b0C. Let the time taken be t\u2082.
\nThe temperature change is \u0394\u03b8\u2082 = 70 – 50 = 20\u00b0C.
\nThe average temperature in this interval is \u03b8_avg\u2082 = (70 + 50) \/ 2 = 60\u00b0C.
\nThe average temperature difference is \u0394T\u2082 = \u03b8_avg\u2082 – \u03b8\u2080 = 60 – 35 = 25\u00b0C.
\nThe average rate of cooling in this interval is approximately Rate\u2082 \u2248 \u0394\u03b8\u2082 \/ t\u2082 = 20\u00b0C \/ t\u2082.
\nBy Newton’s Law of Cooling, Rate\u2082 \u221d \u0394T\u2082, so (20\/t\u2082) \u221d 25.<\/p>\n

Taking the ratio of the rates:
\nRate\u2081 \/ Rate\u2082 = \u0394T\u2081 \/ \u0394T\u2082
\n(4) \/ (20\/t\u2082) = 45 \/ 25
\n4t\u2082 \/ 20 = 9 \/ 5
\nt\u2082 \/ 5 = 9 \/ 5
\nt\u2082 = 9 minutes.
\nThe time taken for the tea to cool from 70\u00b0C to 50\u00b0C is 9 minutes.
\n<\/section>\n","protected":false},"excerpt":{"rendered":"

A cup of tea is placed at a table, and the room temperature is 35\u00b0C. The tea cools from 90\u00b0C to 70\u00b0C in 5 minutes. Which one among the following is the correct time taken for the tea to cool from 70\u00b0C to 50\u00b0C ? [amp_mcq option1=”5 minutes” option2=”7 minutes” option3=”9 minutes” option4=”11 minutes” correct=”option3″] … <\/p>\n

Detailed SolutionA cup of tea is placed at a table, and the room temperature is 35\u00b0C. T<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1091],"tags":[1103,1130,1128],"class_list":["post-87147","post","type-post","status-publish","format-standard","hentry","category-upsc-geoscientist","tag-1103","tag-heat-and-thermodynamics","tag-physics","no-featured-image-padding"],"yoast_head":"\nA cup of tea is placed at a table, and the room temperature is 35\u00b0C. T<\/title>\n<meta name=\"description\" content=\"The correct option is C. According to Newton's Law of Cooling, the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. That is, d\u03b8\/dt \u221d (\u03b8 - \u03b8\u2080), where \u03b8 is the object's temperature and \u03b8\u2080 is the surrounding temperature. For small temperature ranges, the average rate of cooling in an interval can be approximated as proportional to the average temperature difference in that interval.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/exam.pscnotes.com\/mcq\/a-cup-of-tea-is-placed-at-a-table-and-the-room-temperature-is-35c-t\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"A cup of tea is placed at a table, and the room temperature is 35\u00b0C. T\" \/>\n<meta property=\"og:description\" content=\"The correct option is C. According to Newton's Law of Cooling, the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. 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According to Newton's Law of Cooling, the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. That is, d\u03b8\/dt \u221d (\u03b8 - \u03b8\u2080), where \u03b8 is the object's temperature and \u03b8\u2080 is the surrounding temperature. 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