Let T represents temperature of thermometer and S be the constant temperature of oven.
t represents time
Here all temperatures are in °C and time in minute
is proportional to differenc of temperature of thermometer and that of oven.
So = k(T-S), k is constant
=> = k dt
Integrating
= k dt
ln | T-S | = kt + C , C is integration constant
=> ln(S-T) = kt + C as S > T here
When t = 0 , T = 20
So ln(S-20) = k*0 + C
=> C = ln(S-20)
So ln(S-T) = kt + ln(S-20)
=> ln(S-T) - ln(S-20) = kt
=> ln = kt
Two set of values of t, T are as follows
When t = , T = 42
and when t = 1, T = 62
So ln = ..........eq(1)
and ln = k ............eq(2)
Comparing eq(1) and eq(2)
2 ln = ln
=> ln = ln
=> =
=> = (S-62) as S>20, (S - 20)≠0
=> (S-42)2 = (S-20)(S-62)
=> S2 - 84S + 1764 = S2 - 82S + 1240
=> - 84S + 1764 = - 82S + 1240
=> -84S + 82S = 1240 - 1764
=> -2S = -524
=> S = 524/2 = 262
The oven is 262 degree Celsius
Comments