How much heat does it take to bring 0.12 kg of liquid water at 100c to steam at 140c

11. Find the direction of the Force

d1i1m1o1n [39]

Answer:

We answer that before but I forgot

7 0

3 years ago

Setting another object on top of it is this increasing or decreasing

Nostrana [21]

Answer:

Increasing

Explanation:

It’s increasing because your are adding more weight

4 0

3 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

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

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0

3 years ago

What is the kinetic energy of a 1 kg pie if it is thrown at 10 m/s?

nignag [31]

Given

mass (m) = 1 kg

velocity (v) = 10 m/s

kinetic energy ( ke) = ?

,we know

K.E =1/2 m v²

= 1/2 * 1 * 10²

= 100/2

=50 joule

hope it helps :)

7 0

3 years ago

How can geo thermal energy be used ?

Romashka [77]

Heating and Cooling Using Geothermal Heat, Natural Air Conditioning. Another use of this type of energy is natural cooling.Maintenance of Civil Works. Another direct application is to avoid the freezing of bridges, roads, or airports. .Thermal Waters. For the use of heat in hot springs from geothermal energy, deep waters with temperatures between 40 and 150 ° C are used.

7 0

3 years ago

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