All categories

3 months ago

Determine the final state and temperature of 100 g of water originally at 25.0°c after 50.0 kj of heat have been added to it.

1 answer:

4 0

The heat required to raise the temperature of a substance by $\Delta T$ is represented by
$Q=m C_p \Delta T$
where m stands for the mass of the substance and $C_p$ indicates the specific heat of the substance. In this situation, we possess $m=100~g=0.1~Kg$ and $C_p=4.19~KJ/(Kg K)$ , the specific heat of water.
Consequently, we can ascertain the temperature rise $\Delta T$ :
$\Delta T = \frac{Q}{m C_p}= \frac{50~KJ}{0.1~Kg cdot 4.19~KJ/(Kg K)}=119~K =119^{\circ}C$
Initially, the water's temperature was $25^{\circ}C$ , so the end temperature should be
$T_f = 25^{\circ}C+119^{\circ}C=144^{\circ}C$
Thus, the water is expected to be vapor by now.

However, to give a more accurate statement, during the liquid to vapor transition, the heat added to the system is used to break molecular bonds instead of raising the system's temperature. The heat necessary for the phase change from liquid to vapor is expressed as
$Q=m C_L=0.1~Kg \cdot 2265~KJ/Kg=226.5~KJ$
where $C_L$ denotes the latent heat of vaporization for water.
Nevertheless, the initial heat input of 50 KJ is less than this requirement, indicating there isn't sufficient heat to finish the liquid-vapor transition. Therefore, the water will remain in the liquid-vapor change phase at a temperature of $100^{\circ}C$ (the temperature at which the phase change begins)

You might be interested in

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

Maru [3345]

Answer:

The final size is nearly the same as the initial size because the increase in size $1.055\times 10^{- 7}$ is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, $m_{p} = 1.67\times 10^{- 27} kg$

Initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This proton energy is $\simeq 250 GeV$

Thus, the speed of the proton, v $\simeq c$

The time to cover 1 km = 1000 m of distance is calculated as:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

According to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

3 0

2 months ago

Imagine you are a water molecule, carbon atom, or nitrogen atom that is inside a cow. Write a brief explanation of how you got f

Maru [3345]

Answer:

Explanation:

Each of the processes connected to these molecules varies.

For instance, water that has accumulated in the atmosphere returns to the ground as rain. Cows utilize this water from local water sources. This represents one method in which water transitions from the atmosphere to the cow's body.

Regarding carbon and nitrogen, the air inhaled by cows contains nitrogen, oxygen, carbon dioxide, and other gases. These molecules enter the cow through respiration.

6 0

3 months ago

The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of

kicyunya [3294]

3.258 m/s Explanation: The spring constant is assumed to be 263 N/m and the displacement of the spring is also assumed to be 0.7 m; the coefficient of friction between blocks is 0.4. The energy stored in the spring is described by . Given the conservation of energy in the system, the speed of the 8 kg block just prior to collision is 3.258 m/s.

7 0

2 months ago

The force shown in the attached figure is the net eastward force acting on a ball. The force starts rising at t = 0.012 s, falls

ValentinkaMS [3465]

Impulse can be expressed as the integral of F(t) dt from 0.012 s to 0.062 s

Since the function F(t) is unknown, an estimation method will be applied.

The integral represents the area under the curve.

The task suggests approximating this area as a triangle.

For this triangle, the base measures: 0.062 s - 0.012 s = 0.050 s.

The height corresponds to the maximum force of 35 N.

Consequently, the area computes to [1/2] * (0.05 s) * (35 N) = 0.875 N*s.

Final result: 0.875 N*s

6 0

3 months ago