# Tutor profile: Prabath R.

## Questions

### Subject: Linear Algebra

The cake production company want to start packing the cakes after one-hour the cake is baked. The initial temperature of the baked cake is 60℃ and it got 15 minutes to reduce temperature to 50℃.Room temperature is 20℃ and they can do packing after it reaches to 30℃.Can they start the packing process as they desired?

Solution dT/dt=k(T-20) 1/(T-20) dT/dt=k Integrating with respect to time we get, ln〖(T-20)=kt+C〗 T-20=e^(kt+c) T= e^c.e^kt+20 -------------- (1) When t = 0, t = 60 60=e^c+ e^c=40 So that, T=40.e^kt+20 When temperature reduces to 30℃ time should exceed 60 min, If it is they can start packing. When t = 15, T = 50 50=40 e^k(15) +20 e^15k= 30/40 e^k=〖( 3/4)〗^(1/15) Substituting into (1), T=50.e^kt+20 So that let's consider T = 30℃ 30 = 50.〖( 3/4)〗^(t/15)+20 〖( 3/4)〗^(t/15)= 1/5 〖ln( 3/4)〗^(t/15)= ln 1/5 t/15= (ln(0.2))/(ln(0.75)) t=83.92 Answer The 60 min is exceeded, so they can’t start packing.

### Subject: Mechanical Engineering

Project description All necessary logic and calculations Finite element analysis (Forces, thermal) Necessary engineering drawings Possible prototype design General literature parts

All the points will be included in the project report,

### Subject: Differential Equations

The cake production company want to start packing the cakes after one-hour the cake is baked. The initial temperature of the baked cake is 60℃ and it got 15 minutes to reduce temperature to 50℃.Room temperature is 20℃ and they can do packing after it reaches to 30℃.Can they start the packing process as they desired?

Solution dT/dt=k(T-20) 1/(T-20) dT/dt=k Integrating with respect to time we get, ln〖(T-20)=kt+C〗 T-20=e^(kt+c) T= e^c.e^kt+20 -------------- (1) When t = 0, t = 60 60=e^c+ e^c=40 So that, T=40.e^kt+20 When temperature reduces to 30℃ time should exceed 60 min, If it is they can start packing. When t = 15, T = 50 50=40 e^k(15) +20 e^15k= 30/40 e^k=〖( 3/4)〗^(1/15) Substituting into (1), T=50.e^kt+20 So that let's consider T = 30℃ 30 = 50.〖( 3/4)〗^(t/15)+20 〖( 3/4)〗^(t/15)= 1/5 〖ln( 3/4)〗^(t/15)= ln 1/5 t/15= (ln(0.2))/(ln(0.75)) t=83.92 Answer The 60 min is exceeded, so they can’t start packing.

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