Question

A steel tank is completely filled with 2.90m3 of ethanol when the tank and the ethanol are at 33.0 °C. When the tank and its contents have cooled to 20 °C, What additional ethanol can be put into the tank?​

Answer 1

`0.027m^(3)`

Step-by-step explanation:

First, we write down what we know:

`\Delta{V}_(ethanol) = Change\; in\; volume\; of\; the\; ethanol.\\\beta_(ethanol) = 75*{10^(-5)}\;{^\circ{C}^(-1)}\\\beta_(steel) = 36*{10^(-5)}\;{^\circ{C}^(-1)}\\V_(0) = 2.90m^3\\\Delta{T} = T_1-T_0 = 20 - 33 = -13{^\circ{C}}`

In order to solve this problem, we have to calculate the final volumes of the ethanol and steel respectively and then find their difference, which will give us the volume of ethanol that we can add once the container in cooled.

`\Delta{V}_(ethanol)= V_(ethanol)-V_0\\Rearrange\; to\; find\; V_(ethanol)\\ V_(ethanol) = \Delta{V}_(ethanol) +V_0`

`\Delta{V}_(steel)= V_(steel)-V_0\\Rearrange\; to\; find\; V_(steel)\\ V_(steel) = \Delta{V}_(steel) +V_0`

Which gives:

`V_(steel)-V_(ethanol)=volume\;we\;can\;add=V_(free)\\V_(free) = \beta_(steel)V_0\Delta{T}+V_0-(\beta_(ethanol)V_0\Delta{T}+V_0)\\V_(free) = V_0\Delta{T}( \beta_(steel)-\beta_(ethanol))\\V_(free) = (2.9m^3)( -13{^\circ{C}})(36*{10^(-5)}\;{^\circ{C}^(-1)}-75*{10^(-5)}\;{^\circ{C}^(-1)})\\V_(free) = 0.027m^3 = 27l`

Answer 2

V1/T1 = V2/T2 => V2 = V1/T1 * T2 Using the information given, V1 = 1.9 m3 T1 = 32.0 C T2 = 18.0 C we can get: V2 = 1.9/32 * 18 = 1.06875, the volume that would be taken up by the existing ethanol after the temperature change is 1.07. If the capacity is 1.9 m3, then the amount of additional that can be added is 1.90-1.07 which is 0.83 m3.

Step-by-step explanation: