Find the direction of its average acceleration. Find the direction of its average acceleration. 140 ∘ measured counterclockwise from the +x axis 41 ∘ measured counterclockwise from the +x axis 49 ∘ measured counterclockwise from the +x axis 229 ∘ measured counterclockwise from the +x axis

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emily12403

05.01.2021

Physics

The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.

Explanation:

We notice that the student is mixing a substance with a high temperature and another substance with a low temperature, where the first release heat to the latter one until thermal equilibrium is reached. By the First Law of Thermodynamics and assuming that the entire system has no energy interactions with the surroundings, we get the following model:

$\Delta U_{x}+\Delta U_{w} = 0$ (1)

Where $\Delta U_{x}$ and $\Delta U_{w}$ are the changes in internal energy for the unknown substance and water, measured in joules.

By definition of internal energy, we expand the equation above now:

$m_{x}\cdot c_{x}\cdot (T_{o,x}-T_{f,x})+m_{w}\cdot c_{w}\cdot (T_{o,w}-T_{f,w}) = 0$ (2)

Where:

$m_{x}$ , $m_{w}$ - Masses of the unknown substance and water, measured in kilograms.

$c_{x}$ , $c_{w}$ - Specific heats of the unknown substance and water, measured in joules per kilogram-degree Celsius.

$T_{o,x}$ , $T_{f,x}$ - Initial and final temperatures of the unknown substance, measured in degrees Celsius.

$T_{o,w}$ , $T_{f,w}$ - Initial and final temperatures of water, measured in degrees Celsius.

Then, we clear the specific heat of the unknown substance:

$c_{x} = \frac{m_{w}\cdot c_{w}\cdot (T_{f,w}-T_{o,w})}{m_{x}\cdot (T_{o,x}-T_{f,x})}$

If we know that $m_{w} = m_{x} = 0.075\,kg$ , $c_{w} = 4186\,\frac{J}{kg\cdot^{\circ}C}$ , $T_{f,w} = T_{f,x} = 31.15\,^{\circ}C$ , $T_{o,x} = 96.5\,^{\circ}C$ and $T_{o,w} = 25\,^{\circ}C$ , then the heat capacity of the unknown substance is:

$c_{x} = \frac{(0.075\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (31.15\,^{\circ}C-25\,^{\circ}C)}{(0.075\,kg)\cdot (96.5\,^{\circ}C-31.15^{\circ}C)}$

$c_{x} = 393.939\,\frac{J}{kg\cdot ^{\circ}C}$

The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.

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