A pan pizza cools from 91^(@)C " to "79^(@)C in 2 minutes, on a summer day, when the room temperature is 25^(@)C. How long will the pan pizza take to cool from 91^(@)C to 79^(@)C, on a winter day, when the room temperature is 5^(@)C?

A pan pizza cools from 91^(@)C " to "79^(@)C in 2 minutes, on a summer

1.	A pan pizza cools from 91^(@)C " to "79^(@)C in 2 minutes, on a summer day, when the room temperature is 25^(@)C. How long will the pan pizza take to cool from 91^(@)C to 79^(@)C, on a winter day, when the room temperature is 5^(@)C?
Answer» `(3)/(2)` minutes 1 minutes `(1)/(2)` minutes `(1)/(4)` minutes Solution :According to Newton.s LAW of cooling `(T_(1) -T_(2))/(t) = K ((T_(1) + T_(2))/(2) - T_(S))` where `T_(s)` is the surrounding temperature. For the first case, `T_(1) = 91^(@)C, T_(2) = 79^(@)C, T_(s) = 25^(@)C, t=2` min `:. (91^(@)C - 79^(@)C)/(2 "min") = K ((91^(@)C + 79^(@)C)/(2) - 25^(@)C)` or `(12^(@)C)/(2"min") = K(60^(@)C)` ..(i) For the second case, `T_(1) = 91^(@)C, T_(2) = 79^(@)C, T_(s) = 5^(@)C, t=`? `:. (91^(@)C - 79^(@)C)/(t) =K ((91^(@)C + 79^(@)C)/(2) -5^(@)C)` or `(12^(@)C)/(t) = K (80^(@)C)` ...(II) Dividing eqn. (i) by eqn, (ii) we get `(t)/(2"min") = (60^(@)C)/(80^(@)C) or t= (3)/(4) (2"min")= (3)/(2)` min