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`        Hot water cools from 60°c to50°c in first 10 minutes and to 42°c in the next 10minutes the temperature of the surrounding is`
one year ago

Arun
22576 Points
```							Dear Rohit average temp. in first case = (60+50) /2 = 55 Caverage temp. in second case = (50+42) /2 = 46 Cas frm law of coolingdT/dt = -k(T-To )  where To is temp. of surroundingcase 1 : dT = 10 C and dt = 10 minsso,  1=-k(55-To )         ....(1)case 2 : dT = 8 C and dt = 10 minsso,  0.8=-k(46-To )       .........(2)From 1 and 2 , we haveTemp of surr. as 10 CRegardsrun (askIITians forum expert)
```
one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions