How to solve Newton s Law of Cooling when not given initial temperature?

Well, you have 2 unknowns and 2 points. Set up a system of equations, solve for c and k, then find T(0)

415 = 75 + c * e^(-10k)

340 = c * e^(-10k)

347 = 75 + c * e^(-20k)

272 = c * e^(-20k)

340 = c * e^(-10k)

272 = c * e^(-20k)

340 / 272 = c * e^(-10k) / (c * e^(-20k))

340/272 = (c/c) * e^(-10k - (-20k))

4 * 85 / (4 * 68) = 1 * e^(20k - 10k)

85/68 = e^(10k)

ln(85/68) = 10k

(1/10) * ln(85/68) = k

340 = c * e^(-10k)

340 = c * e^(-10 * (1/10) * ln(85/68))

340 = c * e^(-ln(85/68))

340 = c * e^(ln(68/85))

340 = c * (68/85)

340 * 85 / 68 = c

85 * 85 * 4 / 68 = c

85 * 85 / 17 = c

85 * 5 = c

425 = c

Of course, I just realized, while working out the problem, that 85/68 can be further reduced. I'm not too quick on my 17's table sometimes

T(t) = 75 + c * e^(-kt)

c = 425

k = (1/10) * ln(85/68) = (1/10) * ln(5/4)

T(t) = 75 + 425 * e^((-1) * (1/10) * ln(5/4) * t)

T(t) = 75 + 425 * e^((1/10) * ln(4/5) * t)

T(t) = 75 + 425 * (4/5)^(t/10)

T(0) = 75 + 425 * (4/5)^(0/10)

T(0) = 75 + 425 * 1

T(0) = 500

Initial temperature is 500 F

"c" and "k" are constants You have two temperatures (different "t") so you can calculte them using simultaneous equations

415 =75 + ce^(-10k)

and

347 = 75 + ce^(-20k)

solve for c and k