A metal is heated up to a temperature of 500∘C. It is then exposed to a temperature of 38∘C. After 2 minutes, the temperature of the metal becomes 190∘C. When will the temperature be 100∘C?

Name: A metal is heated up to a temperature of 500∘C. It is then exposed to a temperature of 38∘C. After 2 minutes, the temperature of the metal becomes 190∘C. When will the temperature be 100∘C?
Uploaded: 2022-02-23T17:22:57+00:00
Description: A metal is heated up to a temperature of 500∘C. It is then exposed to a temperature of 38∘C. After 2 minutes, the temperature of the metal becomes 190∘C. When will the temperature be 100∘C?

Initial temp =500∘C Ta=Ambient temp=38∘C After 2 min, temp =190∘C Calculate: Time when temp reaches to 100∘C dTdt=−K(T−Ta) dTT−Ta=−K dt ∫500190dTT−Ta=−K∫02dt t=0,T=500∘C t=2,T=190∘C [ln⁡(T−Ta)]500190=−K(2−0) −ln⁡23176=−2K⇒k=12ln⁡23176 ∫dTT−Ta=−K∫dt t=0,T=500∘C T=t,T=100∘C ∫500100dTT−Ta=−k∫0tdt [ln⁡(T−Ta)]500100=−k(t−0) ln⁡(100−38)−ln⁡(500−38)=−kt ln⁡(100−38500−38)=−12ln⁡(23176)t Answer: t=3.61 minute

The mean temperature of the metal as it cools from 500∘C to 190∘C is, θ=500+1902 θ=6902=345∘C The rate of decrease of temperature is, 500−1902=155 deg Cmin According to the Newtons

Ta=Ambient temp=38∘C=190∘C100∘CdTdt=−K(T−Ta)dTT−Ta=−K dt∫500190dTT−Ta=−K∫02dtt=0,T=500∘Ct=2,T=190∘C[ln⁡(T−Ta)]500190=−K(2−0)−ln⁡23176=−2K⇒k=12ln⁡23176∫dTT−Ta=−K∫dtt=0,T=500∘CT=t,T=100∘C∫500100dTT−Ta=−k∫0tdt[ln⁡(T−Ta)]500100=−k(t−0)ln⁡(100−38)−ln⁡(500−38)=−ktln⁡(100−38500−38)=−12ln⁡(23176)tAnswer: t=3.61 minute

Initial temp =500∘C Ta=Ambient temp=38∘C After 2 min, temp =190∘C Calculate: Time when temp reaches to 100∘C dTdt=−K(T−Ta) dTT−Ta=−K dt ∫500190dTT−Ta=−K∫02dt t=0,T=500∘C t=2,T=190∘C [ln⁡(T−Ta)]500190=−K(2−0) −ln⁡23176=−2K⇒k=12ln⁡23176 ∫dTT−Ta=−K∫dt t=0,T=500∘C T=t,T=100∘C ∫500100dTT−Ta=−k∫0tdt [ln⁡(T−Ta)]500100=−k(t−0) ln⁡(100−38)−ln⁡(500−38)=−kt ln⁡(100−38500−38)=−12ln⁡(23176)t Answer: t=3.61 minute

The mean temperature of the metal as it cools from 500∘C to 190∘C is, θ=500+1902 θ=6902=345∘C The rate of decrease of temperature is, 500−1902=155 deg Cmin According to the Newtons

Ta=Ambient temp=38∘C=190∘C100∘CdTdt=−K(T−Ta)dTT−Ta=−K dt∫500190dTT−Ta=−K∫02dtt=0,T=500∘Ct=2,T=190∘C[ln⁡(T−Ta)]500190=−K(2−0)−ln⁡23176=−2K⇒k=12ln⁡23176∫dTT−Ta=−K∫dtt=0,T=500∘CT=t,T=100∘C∫500100dTT−Ta=−k∫0tdt[ln⁡(T−Ta)]500100=−k(t−0)ln⁡(100−38)−ln⁡(500−38)=−ktln⁡(100−38500−38)=−12ln⁡(23176)tAnswer: t=3.61 minute

"A metal is heated up to a temperature..." was answered by students like you

Stefan Hendricks

Answered question

2021-12-29

A metal is heated up to a temperature of

500^{\circ} C

. It is then exposed to a temperature of

38^{\circ} C

. After 2 minutes, the temperature of the metal becomes

190^{\circ} C

. When will the temperature be

100^{\circ} C

Answer & Explanation

puhnut1m

Beginner2021-12-30Added 33 answers

Initial temp

= 500^{\circ} C

T_{a} = Ambient temp = 38^{\circ} C

After 2 min, temp

= 190^{\circ} C

Calculate: Time when temp reaches to

100^{\circ} C

\frac{d T}{d t} = - K (T - T_{a})

\frac{d T}{T - T_{a}} = - K d t

\int_{500}^{190} \frac{d T}{T - T_{a}} = - K \int_{0}^{2} d t

t = 0, T = 500^{\circ} C

t = 2, T = 190^{\circ} C

{[\ln (T - T_{a})]}_{500}^{190} = - K (2 - 0)

- \ln \frac{231}{76} = - 2 K \Rightarrow k = \frac{1}{2} \ln \frac{231}{76}

\int \frac{d T}{T - T_{a}} = - K \int d t

t = 0, T = 500^{\circ} C

T = t, T = 100^{\circ} C

\int_{500}^{100} \frac{d T}{T - T_{a}} = - k \int_{0}^{t} d t

{[\ln (T - T_{a})]}_{500}^{100} = - k (t - 0)

\ln (100 - 38) - \ln (500 - 38) = - k t

\ln (\frac{100 - 38}{500 - 38}) = - \frac{1}{2} \ln (\frac{231}{76}) t

Answer:

t = 3.61 minute

Neunassauk8

Beginner2021-12-31Added 30 answers

The mean temperature of the metal as it cools from

500^{\circ} C to 190^{\circ} C

is,

θ = \frac{500 + 190}{2}

θ = \frac{690}{2} = 345^{\circ} C

The rate of decrease of temperature is,

\frac{500 - 190}{2} = 155 d e g \frac{C}{min}

According to the Newtons

karton

Expert2022-01-09Added 613 answers

$\begin{matrix} T_{a} = Ambient temp = 38^{\circ} C \\ = 190^{\circ} C \\ 100^{\circ} C \\ \frac{d T}{d t} = - K (T - T_{a}) \\ \frac{d T}{T - T_{a}} = - K d t \\ \int_{500}^{190} \frac{d T}{T - T_{a}} = - K \int_{0}^{2} d t \\ t = 0, T = 500^{\circ} C \\ t = 2, T = 190^{\circ} C \\ [\ln (T - T_{a})]_{500}^{190} = - K (2 - 0) \\ - \ln \frac{231}{76} = - 2 K \Rightarrow k = \frac{1}{2} \ln \frac{231}{76} \\ \int \frac{d T}{T - T_{a}} = - K \int d t \\ t = 0, T = 500^{\circ} C \\ T = t, T = 100^{\circ} C \\ \int_{500}^{100} \frac{d T}{T - T_{a}} = - k \int_{0}^{t} d t \\ [\ln (T - T_{a})]_{500}^{100} = - k (t - 0) \\ \ln (100 - 38) - \ln (500 - 38) = - k t \\ \ln (\frac{100 - 38}{500 - 38}) = - \frac{1}{2} \ln (\frac{231}{76}) t \\ A n s w e r : t = 3.61 minute \end{matrix}$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Didn't find what you were looking for?