Question:A block of temperature H=80ºC sits in a room of constant temperature T=20ºC at time t=0. At time t=12, the block has temperature H=50ºC. The rate of change of temperature of the block (dH/dt) is proportional to the temperature difference of the block and room. Find an equation for H in terms of t and the temperature of the block at time t=24.Answer: dH/dt is proportional to H-T, and this rate of change should be negative as the temperature is decreasing. So dH/dt = -k(H-T), where k is a positive constant. Now, using reciprocal rules for derivatives we havedt/dH = -1/k x 1/(H-T). We can integrate both sides with respect to H to findt = -1/k x ln(H-T) + c, where c is a constant of integration. Now rearranging we getln(H-T) = kc - kt, and raising each side to the power of e we getH-T = ekc x e-kt = Ae-kt, for some constant A=ekc. Substituting in the values in the question we seeH=80, t=0 => A=60, andH=50, t=12 => 30=60e-12k => k =ln(2)/12, soH=20+60e-ln(2)t/12, and so when t=24, H=35.So the temperature of the block at t=24 is 35ºC