$\text { Find the time required for an object to cool from } 340 ^ { \circ } \mathrm { F } \text { to } 250 ^ { \circ } \mathrm { F } \text { by evaluating }$ $t = \frac { 10 } { \ln 2 } \int _ { 250 } ^ { 340 } \frac { 1 } { T - 70 } d T$, where $t$ is time in minutes. Round your answer to four decimal places.
A) $6.4346$ minutes
B) $4.4251$ minutes
C) $3.6907$ minutes
D) $4.0598$ minutes
E) $5.8496$ minutes

Question

Quizplus · Accepted Answer

The integral evaluates to $\ln \left| T - 70 \right|$ from 250 to 340. Calculating gives $\ln(270) - \ln(180)$. Simplifying gives $\ln(\frac{270}{180}) = \ln(\frac{3}{2})$. Multiply by $\frac{10}{\ln 2}$ to get the time: $\frac{10}{\ln 2} \cdot \ln(\frac{3}{2}) \approx 5.8496$ minutes.

t=10ln⁡2∫2503401T−70dTt = \frac { 10 } { \ln 2 } \int _ { 250 } ^ { 340 } \frac { 1 } { T - 70 } d Tt=ln210​∫250340​T−701​dT

$t = \frac { 10 } { \ln 2 } \int _ { 250 } ^ { 340 } \frac { 1 } { T - 70 } d T$