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?

Question

puhnut1m · Accepted Answer

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

Neunassauk8 · Accepted Answer

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

karton · Accepted Answer

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