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### $T = T_s + (T_o - T_s)e^{-kt…$T = T_s + (T_o - T_s)e^{-kt}$At 2:00PM$T = 20 + (80 - 20)e^{-kt}T = 20 + 60e^{-kt}$At 2:03PM$42 = 20 + 60e^{-k(3)}e^{-3k} = \frac{11}{30}e^{-k} = \left( \frac{11}{30} \right)^{1/3}$Hence,$T = 20 + 60\left( \frac{11}{30} \right)^{t/3}$t1 minutes after 2:03, the thermometer reads$T = 20 + 60\left( \frac{11}{30} \right)^{{t_1}/3}$When the thermometer is brought back to the room$T = 80 + (T_o - 80)\left( \frac{11}{30} \right)^{t/3}T = 80 + \left[ 20 + 60\left( \frac{11}{30} \right)^{{t_1}/3} - 80 \right] \left( \frac{11}{30} \right)^{t/3}T = 80 + \left[ 60\left( \frac{11}{30} \right)^{{t_1}/3} - 60 \right] \left( \frac{11}{30} \right)^{t/3}T = 80 + 60 \left[ \left( \frac{11}{30} \right)^{{t_1}/3} - 1 \right] \left( \frac{11}{30} \right)^{t/3}$At 2:10PM, t = 7 - t1$71 = 80 + 60 \left[ \left( \frac{11}{30} \right)^{{t_1}/3} - 1 \right] \left( \frac{11}{30} \right)^{(7 - t_1)/3}-9 = 60 \left[ \left( \frac{11}{30} \right)^{t_1/3} - 1 \right] \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}}-9 = 60 \left[ \left( \frac{11}{30} \right)^{7/3} - \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}} \right]\dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{t_1/3}} = \left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}\dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} = \left( \frac{11}{30} \right)^{t_1/3}\left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = \left( \frac{11}{30} \right)^{t_1}\ln \left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = \ln \left( \frac{11}{30} \right)^{t_1}\ln \left[ \dfrac{\left( \frac{11}{30} \right)^{7/3}}{\left( \frac{11}{30} \right)^{7/3} + \dfrac{9}{60}} \right]^3 = t_1 \ln \left( \frac{11}{30} \right)t_1 = 2.8 ~ \text{min}\$

time = 2:05:48.6PM

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