Question

1a) Calculate the increase in volume of 100cm^3 of murcury when its temperature changes from 10°C to 35°C. The volume coefficient of expansion of murcury, B is 0.00018°C^-1

b) Determine the change in volume of a block of cast iron 5.0cm × 10cm × 6.0cm, when the temperature changes from 15°C to 47°C. The coefficient of linear expansion, a for cast iron is 0.000010°C^-1.

c) A glass flask is filled "to the mark" with 50.00cm^3 of murcury at 18°C. If the flask and its contents are heated to 38°C, hoe much murcury will be above the mark? The coefficient of linear expansion a, for glass is 9.0 × 10^-6°C^-1 and the coefficient of volume expansion, B for murcury is 182×10^-6°C^-1.

Take B = 3a glass as a good approximation
i) Determine the change in volume of the murcury

ii) Determine the change in volume of the glass

iii) Calculate how much mercury will be above the mark

Answer 1

Question #80114, Physics / Other

a) Calculate the increase in volume of $100\mathrm{cm}^3$ of mercury when its temperature changes from $10{}^{\circ}\mathrm{C}$ to $35{}^{\circ}\mathrm{C}$. The volume coefficient of expansion of mercury, B is $0.00018{}^{\circ}\mathrm{C}^{\wedge}-1$

Solution

$$\Delta V = \beta \Delta T V = (0.00018)(35 - 10)(100) = 0.45 \mathrm{cm}^3.$$

b) Determine the change in volume of a block of cast iron $5.0\mathrm{cm} \times 10\mathrm{cm} \times 6.0\mathrm{cm}$, when the temperature changes from $15{}^{\circ}\mathrm{C}$ to $47{}^{\circ}\mathrm{C}$. The coefficient of linear expansion, a for cast iron is $0.000010{}^{\circ}\mathrm{C}^{\wedge}-1$.

Solution

$$\Delta V = 3\alpha \Delta T V = 3(0.000010)(47 - 15)(5)(6)(10) = 0.29 \mathrm{cm}^3.$$

c) A glass flask is filled "to the mark" with $50.00\mathrm{cm}^3$ of mercury at $18{}^{\circ}\mathrm{C}$. If the flask and its contents are heated to $38{}^{\circ}\mathrm{C}$, how much mercury will be above the mark? The coefficient of linear expansion a, for glass is $9.0 \times 10^{\wedge}-6{}^{\circ}\mathrm{C}^{\wedge}-1$ and the coefficient of volume expansion, B for mercury is $182 \times 10^{\wedge}-6{}^{\circ}\mathrm{C}^{\wedge}-1$.

Take $B = 3a$ glass as a good approximation

i) Determine the change in volume of the mercury

ii) Determine the change in volume of the glass

iii) Calculate how much mercury will be above the mark

Solution

i)

$$\Delta V_m = \beta \Delta T V = (0.000182)(38 - 18)(50) = 0.182 \mathrm{cm}^3.$$

ii)

$$\Delta V_g = 3\alpha \Delta T V = 3(0.000009)(38 - 18)(50) = 0.027 \mathrm{cm}^3.$$

iii)

$$V = \Delta V_m - \Delta V_g = 0.182 - 0.027 = 0.155 \mathrm{cm}^3.$$

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