The temperature T of a cooling drops at a rate proportional to the difference T-S, where S is the constant temperature of surrounding medium. A thermometer reading `80^@F` was taken outside. Five minutes later, the thermometer read `60^@F`. After another five minutes the reading was `50^@F`. what was the outside temperature?

The temperature T of a cooling object drops at a rate proportional to the difference T-S, where S is the constant temperature of the surrounding medium. Thus, (dT)/(dt)=-k(T-S) , where k(>0) is constant and t is the time. Solve the differntial equation if it is given that T(0)=150.

Newton 's Law of Cooling : According to Newton 's law of cooling the rate of cooling is directly proportional to the temperature difference of body and its surrounding .If body cools from theta_(1) "to" theta_(2) in t seconds then (ms(theta_(1) -theta_(2))/(t)=alpha[(theta_(1)+theta_(2))/(t)- theta_(0)] . Where theta_(0) is the temperature of the surrounding . Is the statement true?

A body takes 10 minutes to cool from 60^@C to 50^@C . If the temperature of surrounding s is 25^@C , then temperature of body after next 10 minutes will be

A particular substance is being cooled by a stream of cold air (temperature of the air is constant and is 5^(@)C ) where rate of cooling is directly proportional to square of difference of temperature of the substance and the air. If the substance is cooled from 40^(@)C to 30^(@)C in 15 min and temperature after 1 hour is T^(@)C, then find the value of [T]//2, where [.] represents the greatest integer function.

For the following cell, Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s) when the concentration of Zn2+ is 10 times the concentration of Cu2+, the expression for ΔG (in J mol–1) is [F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1V]