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Question: A cylindrical container has a cross sectional area of $A_0 = 1cm^2$ at $0^\circ C$. A scale has been...

A cylindrical container has a cross sectional area of A0=1cm2A_0 = 1cm^2 at 0∘C0^\circ C. A scale has been marked on vertical surface of the container which shows correct reading at 0∘C0^\circ C. A liquid is poured in the container. When the liquid and container is heated to 100∘C100^\circ C, the scale shows the height of the liquid as 83.33 cm. The coefficient of volume expansion for the liquid is γ=0.001∘C−1\gamma = 0.001^\circ C^{-1} and the coefficient of linear expansion of the material of cylindrical container is α=0.0005∘C−1\alpha = 0.0005^\circ C^{-1}. A beaker has 300cm3300cm^3 of same liquid at 0∘C0^\circ C. The two liquids are mixed. Find the final temperature of the mixture assuming that heat exchange takes place between the liquids only, and its specific heat capacity is independent of temperature

Answer

70031∘C\frac{700}{31}^\circ C

Explanation

Solution

The problem involves two main parts: first, determining the initial volume of the liquid in the container at 0∘C0^\circ C based on the scale reading at 100∘C100^\circ C and the thermal expansion properties of the liquid and the container; second, calculating the final temperature when this liquid is mixed with a known volume of the same liquid at a different temperature using the principle of calorimetry.

Part 1: Determine the initial volume of the liquid in the container at 0∘C0^\circ C.

Let A0A_0 be the cross-sectional area of the container at 0∘C0^\circ C, and h0h_0 be the actual height of the liquid column at 0∘C0^\circ C. The volume of the liquid at 0∘C0^\circ C is V0=A0h0V_0 = A_0 h_0. Given A0=1cm2A_0 = 1 cm^2. So, V0=h0V_0 = h_0.

When the liquid and container are heated to Tf=100∘CT_f = 100^\circ C, the temperature change is ΔT=100∘C−0∘C=100∘C\Delta T = 100^\circ C - 0^\circ C = 100^\circ C. The cross-sectional area of the container at 100∘C100^\circ C expands to AfA_f. The coefficient of area expansion is βC=2αC\beta_C = 2\alpha_C, where αC\alpha_C is the coefficient of linear expansion of the container material. Af=A0(1+βCΔT)=A0(1+2αCΔT)A_f = A_0 (1 + \beta_C \Delta T) = A_0 (1 + 2\alpha_C \Delta T). Given αC=0.0005∘C−1\alpha_C = 0.0005^\circ C^{-1}. Af=1cm2(1+2×0.0005∘C−1×100∘C)=1(1+0.001×100)=1(1+0.1)=1.1cm2A_f = 1 cm^2 (1 + 2 \times 0.0005^\circ C^{-1} \times 100^\circ C) = 1 (1 + 0.001 \times 100) = 1 (1 + 0.1) = 1.1 cm^2.

The volume of the liquid at 100∘C100^\circ C expands to VfV_f. The coefficient of volume expansion of the liquid is γL\gamma_L. Vf=V0(1+γLΔT)V_f = V_0 (1 + \gamma_L \Delta T). Given γL=0.001∘C−1\gamma_L = 0.001^\circ C^{-1}. Vf=V0(1+0.001∘C−1×100∘C)=V0(1+0.1)=1.1V0V_f = V_0 (1 + 0.001^\circ C^{-1} \times 100^\circ C) = V_0 (1 + 0.1) = 1.1 V_0.

At 100∘C100^\circ C, the liquid fills the container up to an actual height hfh_f. The volume of the liquid at 100∘C100^\circ C is also Vf=AfhfV_f = A_f h_f. So, 1.1V0=1.1hf1.1 V_0 = 1.1 h_f. Since V0=A0h0=1×h0=h0V_0 = A_0 h_0 = 1 \times h_0 = h_0, we have 1.1h0=1.1hf1.1 h_0 = 1.1 h_f, which implies hf=h0h_f = h_0. The actual height of the liquid column remains the same at 100∘C100^\circ C as it was at 0∘C0^\circ C. This is because γL=2αC\gamma_L = 2\alpha_C, which is the condition for the apparent expansion of the liquid level in a container with expanding area to be zero if the scale did not expand. However, the scale does expand.

The scale is marked at 0∘C0^\circ C. A length L0L_0 at 0∘C0^\circ C becomes Lf=L0(1+αCΔT)L_f = L_0 (1 + \alpha_C \Delta T) at 100∘C100^\circ C. The scale shows a reading of hscale=83.33cmh_{scale} = 83.33 cm at 100∘C100^\circ C. This reading corresponds to a length h0′h_0' at 0∘C0^\circ C such that the actual height hfh_f is the expanded length of h0′h_0'. So, hf=h0′(1+αCΔT)h_f = h_0' (1 + \alpha_C \Delta T). The scale reading hscaleh_{scale} is this length h0′h_0'. hscale=h0′=hf1+αCΔTh_{scale} = h_0' = \frac{h_f}{1 + \alpha_C \Delta T}. Since hf=h0h_f = h_0, we have hscale=h01+αCΔTh_{scale} = \frac{h_0}{1 + \alpha_C \Delta T}. Given hscale=83.33cm=2503cmh_{scale} = 83.33 cm = \frac{250}{3} cm. h0=hscale(1+αCΔT)=2503cm×(1+0.0005∘C−1×100∘C)h_0 = h_{scale} (1 + \alpha_C \Delta T) = \frac{250}{3} cm \times (1 + 0.0005^\circ C^{-1} \times 100^\circ C). h0=2503(1+0.05)=2503×1.05=2503×105100=250×105300=25×10530=5×1056=5×352=1752=87.5cmh_0 = \frac{250}{3} (1 + 0.05) = \frac{250}{3} \times 1.05 = \frac{250}{3} \times \frac{105}{100} = \frac{250 \times 105}{300} = \frac{25 \times 105}{30} = \frac{5 \times 105}{6} = \frac{5 \times 35}{2} = \frac{175}{2} = 87.5 cm. The actual height of the liquid at 0∘C0^\circ C is h0=87.5cmh_0 = 87.5 cm. The volume of the liquid in the container at 0∘C0^\circ C is V0,container=A0h0=1cm2×87.5cm=87.5cm3V_{0,container} = A_0 h_0 = 1 cm^2 \times 87.5 cm = 87.5 cm^3. This liquid is currently at 100∘C100^\circ C.

Part 2: Mixing of liquids.

We have two quantities of the same liquid:

  1. Liquid from the container: Initial temperature T1=100∘CT_1 = 100^\circ C. Volume at 0∘C0^\circ C is V0,1=87.5cm3V_{0,1} = 87.5 cm^3.
  2. Liquid from the beaker: Initial temperature T2=0∘CT_2 = 0^\circ C. Volume at 0∘C0^\circ C is V0,2=300cm3V_{0,2} = 300 cm^3.

Let ρ0\rho_0 be the density of the liquid at 0∘C0^\circ C. The masses of the two quantities of liquid are m1=ρ0V0,1m_1 = \rho_0 V_{0,1} and m2=ρ0V0,2m_2 = \rho_0 V_{0,2}. m1=ρ0×87.5m_1 = \rho_0 \times 87.5. m2=ρ0×300m_2 = \rho_0 \times 300.

Let cc be the specific heat capacity of the liquid, which is assumed to be independent of temperature. When the two liquids are mixed, assuming no heat exchange with the surroundings, the heat lost by the hotter liquid equals the heat gained by the colder liquid. Let TfT_f be the final temperature of the mixture. Heat lost by liquid 1 = m1c(T1−Tf)m_1 c (T_1 - T_f). Heat gained by liquid 2 = m2c(Tf−T2)m_2 c (T_f - T_2). m1c(T1−Tf)=m2c(Tf−T2)m_1 c (T_1 - T_f) = m_2 c (T_f - T_2). (ρ0×87.5)c(100−Tf)=(ρ0×300)c(Tf−0)(\rho_0 \times 87.5) c (100 - T_f) = (\rho_0 \times 300) c (T_f - 0). Assuming ρ0≠0\rho_0 \neq 0 and c≠0c \neq 0, we can cancel them out. 87.5(100−Tf)=300Tf87.5 (100 - T_f) = 300 T_f. 8750−87.5Tf=300Tf8750 - 87.5 T_f = 300 T_f. 8750=300Tf+87.5Tf8750 = 300 T_f + 87.5 T_f. 8750=387.5Tf8750 = 387.5 T_f. Tf=8750387.5T_f = \frac{8750}{387.5}. 387.5=7752387.5 = \frac{775}{2}. Tf=87507752=8750×2775T_f = \frac{8750}{\frac{775}{2}} = \frac{8750 \times 2}{775}. 8750=875×108750 = 875 \times 10. 775=25×31775 = 25 \times 31. 875=25×35875 = 25 \times 35. Tf=(25×35)×10×225×31=35×10×231=70031T_f = \frac{(25 \times 35) \times 10 \times 2}{25 \times 31} = \frac{35 \times 10 \times 2}{31} = \frac{700}{31}. Tf=70031≈22.58∘CT_f = \frac{700}{31} \approx 22.58^\circ C.

The final temperature of the mixture is 70031∘C\frac{700}{31}^\circ C.

Explanation of the solution:

  1. Calculate the actual height h0h_0 of the liquid at 0∘C0^\circ C using the scale reading hscaleh_{scale} at 100∘C100^\circ C and the thermal expansion of the container scale: h0=hscale(1+αCΔT)h_0 = h_{scale} (1 + \alpha_C \Delta T).
  2. Calculate the volume of the liquid in the container at 0∘C0^\circ C: V0,container=A0h0V_{0,container} = A_0 h_0. This liquid is at 100∘C100^\circ C when mixed.
  3. Identify the second liquid: 300cm3300 cm^3 at 0∘C0^\circ C. This is its volume at 0∘C0^\circ C, V0,beaker=300cm3V_{0,beaker} = 300 cm^3.
  4. Use the principle of calorimetry: Heat lost by the hot liquid equals the heat gained by the cold liquid. m1c(T1−Tf)=m2c(Tf−T2)m_1 c (T_1 - T_f) = m_2 c (T_f - T_2).
  5. Express masses in terms of volumes at 0∘C0^\circ C and density at 0∘C0^\circ C: m=ρ0V0m = \rho_0 V_0.
  6. Substitute values and solve for the final temperature TfT_f.