A body initially at 80^(@)C cools to 64^(@)C in 5 minutes and to 52^(@) C in 10 minutes.What will be the temprature after 15 minutes and what is the temperature of the surroundings ?

A body initially at 80^(@)C cools to 64^(@)C in 5 minutes and to 52^(@

1.	A body initially at 80^(@)C cools to 64^(@)C in 5 minutes and to 52^(@) C in 10 minutes.What will be the temprature after 15 minutes and what is the temperature of the surroundings ?
Answer» Solution :`(80-64)//5alpha(80+64)//2-(theta_(0)), (64-52)//5alpha(64+52)//2-(theta_(0))` solving `theta_(0)=16^(@)C, (52-THETA)//5alpha[(52+theta)//2]-16`solvin`theta=43^(@)C`