a) temperature after 58 minutes
= 21.2(1.0418)^58 = 227.94°
b)
temperature after 1.6 hrs or 96 minutes
= 227
94(.9846)^96 = 51.38°
c) to go from 227.94 back to 21.2 ...
227.94(.9846)^t = 21.2
.9846^t = .093007
t log .9846 = log .093007
t = log .093007/log .9846 = 153.035 minutes
= appr 2 hrs. 33 minutes
1. A steel plate initially at 21.2°C is placed in a furnace at 785°C for 58.0 minutes. While in the furnace the temperature of the plate increases at a rate of 4.18% per minute. After 58.0 minutes, the plate is removed from the furnace and placed on a concrete slab to cool. While on the slab, it cools exponentially at a rate of 1.54% per minute.
a) Calculate the temperature of the steel plate immediately after being removed from the furnace.
b) Determine the temperature of the steel plate 1.60 hours after being removed from the furnace.
c) How long will it take the plate to cool to its initial temperature?
1 answer