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Mkey [24]
3 years ago
5

147.5 ml of ethyl alcohol (coefficient of volume expansion = 1120 x 10-6K-1) at a temperature of 273.1 K is measured into a 150.

0 ml glass beaker (whose coefficient of volumeexpansion is negligible by comparison). To what temperature does the beaker have to be warmed for it to be completely full (i.e. for the volume of the alcohol to reach 150.0 ml)?
Physics
1 answer:
marishachu [46]3 years ago
8 0

Answer:

288.2 K

Explanation:

\beta = Volumetric expansion coefficient = 1120\times 10^{-6}\ /K

T_i = Initial temperature = 273.1 K

T_f = Final temperature

v_0 = Original volume = 150 mL

Change in volume is given by

\Delta v=\beta v_0(T_f-T_i)\\\Rightarrow T_f=\frac{\Delta v}{\beta v_0}+T_i\\\Rightarrow T_f=\frac{150-147.5}{1120\times 10^{-6}\times 147.5}+273.1\\\Rightarrow T_f=288.2\ K

The temperature of the ethyl alcohol should be 288.2 K to reach 150 mL

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How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r
Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}

The heat (Q) can be calculated using the following expression:

Q=c \times m \times \Delta T

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

162mL.\frac{0.997g}{mL} = 162g

Then,

Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g

The volume of water is:

139g.\frac{1mL}{0.997g} =139mL

7 0
3 years ago
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