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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

A projectile is fired from a height of 80 M above sea level, horizontally with a speed of 360 M / S, calculate: The time it take

Maslowich

Answer:

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

$x = x_{o}+v_{o,x} \cdot t$

$y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$x_{o}$ , $y_{o}$ - Initial horizontal and vertical position of the projectile, measured in meters.

$v_{o,x}$ , $v_{o,y}$ - Initial horizontal and vertical speed of the projectile, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

$x$ , $y$ - Current horizontal and vertical position of the projectile, measured in meters.

Given that $x_{o} = 0\,m$ , $y_{o} = 80\,m$ , $v_{o,x} = 360\,\frac{m}{s}$ , $v_{o,y} = 0\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the kinematic equations are, respectively:

$x = 360\cdot t$

$y = 80-4.094\cdot t^{2}$

(a) If $y = 0\,m$ , the time taken for the projectile to reach the water is:

$80 - 4.094\cdot t^{2} = 0$

$t = \sqrt{\frac{80}{4.094} }\,s$

$t \approx 4.420\,s$

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If $t \approx 4.420\,s$ , the horizontal scope of the projectile is:

$x = 360\cdot (4.420)$

$x = 1591.2\,m$

The horizontal scope of the projectile is 1591.2 meters.

(c) If $t = 2\,s$ , the height that remains to descend is:

$y = 80-4.094\cdot (2)^{2}$

$y = 63.624\,m$

The remaining height to descend after 2 seconds of being launched is 63.624 meters.

6 0

3 years ago

Bill and Nageen have each built an electric motor and they are testing them by having them lift boxes. Bill's motor lifts a box

photoshop1234 [79]

Answer:

Bill's motor power: W_B = F x S / T = F x 0.35 / 2= 0.175F

Nageen's motor power: W_N = F x S / T = F x 0.35 / 1.8 = 0.194F

=> 0.194F > 0.175F => Nageen's motor applied more power to the box than Bill's motor.

5 0

3 years ago

One example of a very complex land ecosystem full of thousands of kinds of living things

german

A Tropical Rain Forest. It has both Biotic and Abiotic factors which will contribute to thousands of living things.

6 0

3 years ago

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How is the kinetic energy of the particles that make up a substance different

never [62]

Answer:

Explanation:

The faster particles move, the more kinetic energy they have. Within a given substance, such as water, particles in the solid state have the least amount of kinetic energy. This is because they only vibrate in place. Particles in the liquid state move faster than particles in the solid state.

3 0

4 years ago

Read 2 more answers