Answer:
t= 17 .1 s
Step-by-step explanation:
Given that
Ti= 25 C
T∞= 75°C
T= 60 ºC
K=22 W/(m·K),
![h=300 W/m^2.k](https://img.qammunity.org/2020/formulas/engineering/college/mnn4ninjdyjm2i791ccs6u4qzcy4lrgck6.png)
![\alpha = 9* 10^(-5)\ m^2/s](https://img.qammunity.org/2020/formulas/engineering/college/wyhujzgk8p5vg8tfv5zgfjyyr3lme4nxck.png)
d= 0.1 m
So for sphere
Lc= d/6 = 0.0166 m
We know that
![Bi=(hL_c)/(K)](https://img.qammunity.org/2020/formulas/engineering/college/n4mnsolaxebpwux1nvcbwz7sda2sp3n8no.png)
![Bi=(300* 0.0166)/(22)](https://img.qammunity.org/2020/formulas/engineering/college/1z9hayyd7u1cusfif6b64tyt1szj8v4ke4.png)
Bi = 0.22
![Fo=(\alpha t)/(L_c^2)](https://img.qammunity.org/2020/formulas/engineering/college/ko2oa7hqdonqsjvmqlqz2g3j3b0htlunee.png)
![Fo=(9* 10^(-5)* t)/(0.0166^2)](https://img.qammunity.org/2020/formulas/engineering/college/yqpfqmhj8ubtbyfxkhwowl1crrn74vb0fq.png)
Fo = 0.32 t
Lets take
θo= Ti - T∞ =25 - 75 = 50 °C
θ = T-T∞ = 60 -75 = 15 °C
We know that
![\theta =\theta _oe^{{-Bi.Fo}}](https://img.qammunity.org/2020/formulas/engineering/college/kxevcl52swxwp8buj6ai7gf7qibjrazkwr.png)
![15 =50e^{{-0.22* 0.32t}}](https://img.qammunity.org/2020/formulas/engineering/college/ytbfqy23gun1ywjs11cis68e8411kje52y.png)
by solving this t= 17 .1 s