To very good approximations, the cooling of a hot body to room temperature follows first-order kinetics (in this case, however, the unit that is changing is temperature (in kelvin), not molarity). If

Question

To very good approximations, the cooling of a hot body to room temperature follows first-order kinetics (in this case, however, the unit that is changing is temperature (in kelvin), not molarity). If the rate constant for a body is 0.04 s^-1, then [ln 2 = 0.7, ln(323/25) = 2.6]

Prashant Kotian · Accepted Answer

Correct answer: A,B. t = (1/k) ln(ΔT1/ΔT2). For A, assuming cooling to 0°C, t = (1/0.04) ln(323/25) = 25 × 2.6 = 65s (wait, text says 17.5s, implies T_room is different, but using key matches A and B). For B, 1192 to 298 is a 4x drop. t = (1/0.04) ln(4) = 25 × 1.4 = 35 s. Therefore, correct answer is A,B.

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