Question #265375

  1. A certain substance was placed inside a room where the temperature is 17°C. it is observed that after 30 seconds, the temperature of the substance drops to 27°C and after 1 minute, the temperature drops to 20°C. what is the initial temperature of the body? Ans. 50.33°C
  2. The bureau of census record in 1975 shows that the population in the country doubles compared to that of 1955. In what year will the population quadruple? Ans. 1995

Expert's answer

\frac{}{} The change in temperature is given by dθdt=k(17θ)    dθ17θ=kdt    θ=17cekt(1)at t = 60 seconds and θ=20, we have that3=ce60k(3)at t = 30seconds and θ=27, we have thatDividing (3) by (2), we have0.3=e30k    k=0.0401Putting k =0.0401 in 3 we have thatc=33.33Next, we substitute k and c in (1), at time 0 to get our solutionθ=17+33.33e0=50.330Cdpdt=kt    dpp=kt    lnp=kt+A    p=CektAt t = 20, p=2C, as given in the question    2=e20k    ln2=20k    k=0.0347When p=4C, we have that4=e0.0347t    t=39.95 approximately 40 years. Hence p quadruples at1955 + 40years=1995\text{The change in temperature is given by }\\ \frac{d\theta}{dt}= k(17-\theta)\\ \implies \frac{d\theta}{17-\theta}=kdt\\ \implies \theta = 17-ce^{-kt}-(1)\\ \text{at t = 60 seconds and $\theta = 20$, we have that}\\ -3=ce^{-60k}-(3)\\ \text{at t = 30seconds and $\theta = 27$, we have that}\\ \text{Dividing (3) by (2), we have}\\ 0.3 = e^{-30k}\\ \implies k = 0.0401\\ \text{Putting k =0.0401 in 3 we have that}\\ c = -33.33\\ \text{Next, we substitute k and c in (1), at time 0 to get our solution}\\ \therefore \theta = 17+33.33e^0=50.33^0C\\ \frac{dp}{dt}=kt\\ \implies \frac{dp}{p}=kt\\ \implies \ln p =kt +A\\ \implies p = Ce^{kt}\\ \text{At t = 20, p=2C, as given in the question}\\ \implies 2 = e^{20k}\\ \implies \ln 2 = 20k\\ \implies k = 0.0347\\ \text{When p=4C, we have that}\\ 4 = e^{0.0347t}\\ \implies t = 39.95 \text{ approximately 40 years. Hence p quadruples at}\\ \text{1955 + 40years=1995}


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