The change in temperature is given by dtdθ=k(17−θ)⟹17−θdθ=kdt⟹θ=17−ce−kt−(1)at t = 60 seconds and θ=20, we have that−3=ce−60k−(3)at t = 30seconds and θ=27, we have thatDividing (3) by (2), we have0.3=e−30k⟹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 solution∴θ=17+33.33e0=50.330Cdtdp=kt⟹pdp=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
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