A thermometer is taken from a room where the temperature is 21C to the outdoors,

Question

A thermometer is taken from a room where the temperature is 21C to the outdoors, where the temperature is -12C . After one minute the thermometer reads 6C. (a) What will the reading on the thermometer be after 4 more minutes? (b) When will the thermometer read -11C ?

Explanation / Answer

Solve in the same approach.... A thermometer is taken from a room where the temperature is 20°C to the outdoors, where the temperature is 7°C. After one minute the thermometer reads 13°C. (Round the answers to one decimal place.) (a) What will the reading on the thermometer be after one more minute? .°C (b) When will the thermometer read 8°C? minutes Solution: Use Newton's Law of Cooling. T = T_s + (T_0 - T_s)*e^(-kt) where T = temperature at any instant T_s = temperature of surroundings T_0 = original temperature t = elapsed time k = constant Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7° C envorniment. T = 13, T_0 = 20, T_s = 7, t = 1, k = ? T = T_s + (T_0 - T_s)*e^(-kt) ==> 13 = 7 + (20 - 7)*e^(-k) ==> 13 = 7 + 13*e^(-k) ==> 6 = 13*e^(-k) ==> 6/13 = e^(-k) ==> -k = ln(6/13) ==> k = -ln(6/13) ˜ 0.773 Now that we got out constant, let's calculate some temperatures! T = ?, T_0 = 20, T_s = 7, k = 0.773, t = 2 T = T_s + (T_0 - T_s)*e^(-kt) ==> T = 7 + (20 - 7)*e^[ -(0.773)(2) ] ==> T = 7 + 13*e^(-1.546) ==> T ˜ 9.77° C Therefore, the temperature after one more minute would be about 9.8° C. 2) T = 8, T_0 = 20, T_s = 7, k = 0.773, t = ? T = T_s + (T_0 - T_s)*e^(-kt) ==> 8 = 7 + (20 - 7)*e^(-0.773t) ==> 8 = 7 + 13*e^(-0.773t) ==> 1 = 13*e^(-0.773t) ==> e^(-0.773t) = 1/13 ==> -0.773t = ln(1/13) ==> t = -ln(1/13) / 0.773 ˜ 3.318 minutes Therefore, after about 3.318 minutes after it was put in the 7° C envorniment it would be 8° C.

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