Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

The cavity within a copper [β = 51 × 10-6 (C°)-1] sphere has a volume of 1.180 × 10-3 m3. Into this cavity is placed 1.100 × 10-3 m3 of benzene [β = 1240 × 10-6 (C°)-1]. Both the copper and the benzene have the same temperature. By what amount ΔT should the temperature of the sphere and the benzene within it be increased, so that the liquid just begins to spill out?

The Cavity Within A Copper Β 51 106 C1 Sphere Has A Volume Of 1180 103 M3 Into This Cavity Is Placed 1100 103 M3 Of Benzene Β 1240 106 C1 Both The Copper And Th class=

Sagot :

Answer:

The answer is "[tex]60.74^{\circ}[/tex]".

Explanation:

Cavity and benzene should be extended in equal quantities.

[tex]\to 1.18 \times 10^{-3}\times (1+ \Delta T \times 0.000051) = 1.1\times 10^{-3} \times (1+ \Delta T \times 0.00124)\\\\\to (\frac{1.18}{1.1})\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072\times (1+ \Delta T \times 0.000051) = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 = 1+ \Delta T \times 0.00124\\\\ \to 1.072+ \Delta T \times 0.000054672 - 1- \Delta T \times 0.00124=0\\\\[/tex]

[tex]\to 0.072+ \Delta T \times 0.000054672 - \Delta T \times 0.00124=0\\\\ \to 0.072+ \Delta T ( 0.000054672 -0.00124)=0\\\\ \to \Delta T ( 0.000054672 -0.00124)= -0.072\\\\ \to \Delta T = -\frac{0.072}{( 0.000054672 -0.00124)}\\\\ \to \Delta T = -\frac{0.072}{-0.001185328 }\\[/tex]

[tex]\to \Delta T = \frac{0.072}{0.001185328 }\\\\ \to \Delta T = 60.74^{\circ}\\[/tex]