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OleMash [197]
3 years ago
10

the specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change

?
Physics
1 answer:
defon3 years ago
7 0
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=mC_s \Delta T
where
m is the mass of the substance
C_s the specific heat capacity
\Delta T the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is C_s = 4.2 J/g ^{\circ}C and the amount of heat supplied is Q=31500 J, so if we re-arrange the previous formula we find the increase in temperature of the water:
\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C
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A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located
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<h3>What is Columb's law?</h3>

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Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

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F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241  \times 10^{-19 } \ C

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

#SPJ2

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