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3 years ago

Hi.

Physics

2 answers:

Pani-rosa [81]3 years ago

6 0

<h2>Answer:</h2>

Options a and e are units which are C and T.

While options b, c ,d and f are variables or quantities.

<h3>Explanation:</h3>

The units are the signs or representation of a standard amount of an physical quantity like meter for length.

Here C and T are units because they are not in cursive form.

C is unit of charge Coulomb while T is Tesla the unit of magnetic field strength.

Variables are written in italics or cursive form. As:

<em>c = </em>any constant.
<em>C = c</em>urie constant.
<em>T= </em>Temperature.
<em>t = </em>Time.

zhenek [66]3 years ago

4 0

Answer: Units: C, T and Variables: <em>c, C, t, T</em>

(a) C is the unit of charge Coulomb, (e) T is the unit of magnetic field strength Tesla.

In physics, units are written straight and variables are written in italics. Various italic symbols are used for various quantities like <em>t </em>for time, <em>T</em> for temperature, <em>c</em> for any constant and <em>C</em> for curie constant.

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You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$