F.4 Maths

Original water temp. = 100.
Room temp = 20. Therefore, original difference in temp = 100-20 = 80.
After 1 minute, temp. difference = 80(1-4%) = 80 x 0.96.
After 2 minutes, temp difference = 80 x 0.96 x (1-4%) = 80 x 0.96 x 0.96 = 80 x (0.96)^2.
After 3 minutes, temp difference = 80 x 0.96 x 0.96 x (1-4%) = 80 x 0.96 x 0.96 x 0.96 = 80 x (0.96)^3. Therefore, after t minutes,
Temp. Difference = 80 x (0.96)^t.
b)When t=14. Temp. difference = 80 x (0.96)^14 = 45.2
Therefore, temp. of hot water = 45.2+ room temp. = 45.2 + 20 = 65.
c)When hot water temp = 30. Temp. difference = 30 -20 =10.
Therefore, 10= 80 x(0.96)^t
0.125 = (0.96)^t
t = log 0.125/log 0.96 = 51 min.

b)66degree C

b)65 degree C