Question

A small metal bar, whose initial temperature was 10° C, is dropped into a large container of boiling water. How long will it take the bar to reach 70° C if it is known that its temperature increases 2° during the first second? (The boiling temperature for water is 100° C. Round your answer to one decimal place.) sec How long will it take the bar to reach 98° C? (Round your answer to one decimal place.)

Answer 1

Answer

According to newton law of cooling

$\dfrac{d T}{dt} = k (T - T_a)$

and

$T = Ce^{kt} + T_a$

now At Ta = 100

T₀ = 10 °C

T₁ = 12° C

$T(0) = Ce^{0} + T_a$

10 = C + 100

C = -90

now,

$12 = -90e^k + 100$

$e^k = \dfrac{88}{90}$

$k = ln{\dfrac{88}{90}}$

at time equal to t

T(t) = 70

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$70 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{3}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{3})$

t = 48.88 s

hence, after 48.88 s the temperature of the body will be 70°C

b) time taken to reach 98°C

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$98 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{45}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{45})$

t = 390 s

hence, after 390 s the temperature of the body will be 98°C