Newton's Law of Cooling, but not give t=0 temperature?

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I encountered this Newton's Law of Cooling questions in my textbook and usually it gives the temperature when t = 0, but this time it didn't, which is giving me a hard time.

It says exactly as follows,

A hot object is placed in a room whose temperature is 80F. After 1 minute, the temperature of the object is 160F and its rate of change of temperature is 22F per minute. Find the initial temperature of the object.

Here is my attempt,

dT/dt = k (t - 80)

dT(1/(T-80) = dt k

ln (T-80) = kt +c

T = ce^(kt) + 80

T prime = [ce^(tk)]/k

So, it says

T(1) = 160 F

T prime(1) = 22 F/min

Which gives me,

160 = ce^k + 80

22 = (ce^k)/k

I solved for c in the second eq.

22k/e^k = c

Plug in the first eq.

160 = (22k/e^k) * e^k + 80

80 = 22k

k = 80/22

So I have k now, now I solve for c,

160 = ce^(80/22) + 80

c = 80/e^(80/22)

This all seems correct to me but, BUT when I test it out on my calculator, the values it gives seems way too large.

The textbook has no solution for even numbers, so I am unable to compare answers.

Can you guys spot my mistake...?

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Check your T prime!

(Sorry, this would work better as a comment, but I need one upvote for that)