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2 years ago

Find the specific heat of a substance that requires 8000 J of energy to heat up 400g by 20 C?

Physics

1 answer:

monitta2 years ago

7 0

Answer:

$c=1\ J/g^\circ C$

Explanation:

Given that,

Heat required, Q = 8000 J

Mass, m = 400 g

The change in temperature, $\Delta T = 20^{\circ}$

The heat required due to change in temperature is given by :

$Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{8000 }{400\times 20}\\\\c=1\ J/g^\circ C$

So, the specific heat of the substance is $1\ J/g^\circ C$

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An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

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3 years ago

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2 years ago

An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t

Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

$E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta$

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

$\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0$