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 ...

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

CB
Answered by Connie B. Maths tutor

2856 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do we differentiate y=a^x when 'a' is an non zero real number


How do I differentiate an algebraic expression? (e.g. y=3x^4 - 8x^3 - 3) [the ^ represents x being raised to a power]


Find the equation of the tangent to the circle x^2 + y^2 + 10x + 2y + 13 = 0 at the point (-3, 2)


How do you find the equation of a line at a given point that is tangent to a circle?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning