A cup of coffee take a time ' t' to cool from \(90^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) in a surrounding of…

Explanation

Newton’s Law of Cooling states:

\(\frac{{T}_{1}−{T}_{2}}{t}=k(\frac{{T}_{1}+{T}_{2}}{2}−{T}_{s})\) First Case: Cooling from \(9{0}^{∘}C\)90∘C to \(8{0}^{∘}C\)80∘C

Given:

• Initial Temperature: \({T}_{1}=9{0}^{∘}C\)
• Final Temperature: \({T}_{2}=8{0}^{∘}C\)
• Surrounding Temperature: \({T}_{s}=2{0}^{∘}C\)

Substituting the values into the equation:

\(\frac{90−80}{t}=k(\frac{90+80}{2}−20)\) \(\frac{10}{t}=k(85−20)\) \(\frac{10}{t}=k(65)\)

Solving for \(k\):

\(k=\frac{2}{13t}\)​ Second Case: Cooling from \(8{0}^{∘}C\) to \(6{0}^{∘}C\)

Given:

• Initial Temperature: \({T}_{1}=8{0}^{∘}C\)
• Final Temperature: \({T}_{2}=6{0}^{∘}C\)
• Surrounding Temperature: \({T}_{s}=2{0}^{∘}C\)

Using Newton’s Law:

\(\frac{80−60}{{t}^{′}}=k(\frac{80+60}{2}−20)\) \(\frac{20}{{t}^{′}}=k(50)\)

Substituting \(k=\frac{2}{13t}\)​ from the first case:

\(\frac{20}{{t}^{′}}=\frac{2}{13t}\times 50\) \({t}^{′}=\frac{13t}{5}\)​ Final Answer: \(\frac{13t}{5}\)