MHT CET · Maths · Differential Equations
Water at \(100^{\circ} \mathrm{C}\) cools in 10 minutes to \(80^{\circ} \mathrm{C}\) in a room temperature of \(25^{\circ} \mathrm{C}\), then the temperature of water after 20 minutes will be
- A \(65.33^{\circ} \mathrm{C}\)
- B \(69.33^{\circ} \mathrm{C}\)
- C \(60.33^{\circ} \mathrm{C}\)
- D \(63.33^{\circ} \mathrm{C}\)
Answer & Solution
Correct Answer
(A) \(65.33^{\circ} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\( \frac{\mathrm{d} T}{\mathrm{~d} t}=-k(T-25) \)
\( \Rightarrow T-25=e^{-k t+c} \)
\( \Rightarrow T=25+e^c \cdot e^{-k t} \)
\( \text { for } t=0, T=100 \)
\( \Rightarrow e^c=75 \)
\( \Rightarrow T=25+75 e^{-k t} \)
\( \text { for } t=10, T=80 \)
\( \Rightarrow 80=25+75 e^{-k \times 10} \)
\( \Rightarrow-10 k=\log \left(\frac{11}{55}\right) \)
\( \text {i.e., } T=25+75 e^{\left(\frac{1}{10} \log \frac{11}{15}\right) t} \)
\( \text {Now for } t=20, T=25+75 e^{20 \times \frac{1}{10} \log \frac{11}{15}}=25~+\) \(75 \times\left(\frac{11}{15}\right)^2 \)
\( \Rightarrow T=25+40.33=65.33\)
\( \Rightarrow T-25=e^{-k t+c} \)
\( \Rightarrow T=25+e^c \cdot e^{-k t} \)
\( \text { for } t=0, T=100 \)
\( \Rightarrow e^c=75 \)
\( \Rightarrow T=25+75 e^{-k t} \)
\( \text { for } t=10, T=80 \)
\( \Rightarrow 80=25+75 e^{-k \times 10} \)
\( \Rightarrow-10 k=\log \left(\frac{11}{55}\right) \)
\( \text {i.e., } T=25+75 e^{\left(\frac{1}{10} \log \frac{11}{15}\right) t} \)
\( \text {Now for } t=20, T=25+75 e^{20 \times \frac{1}{10} \log \frac{11}{15}}=25~+\) \(75 \times\left(\frac{11}{15}\right)^2 \)
\( \Rightarrow T=25+40.33=65.33\)
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