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2 years ago

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A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b

oth substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.)

1 answer:

lakkis [162]2 years ago

6 0

Answer:

$T = 17.26 ^oC$

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

$Q_1 = Q_2$

$m_1s_1\Delta T_1 = m_2s_2\Delta T_2$

so we will have

$32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)$

so we have

$(45.8 - T) = 15.1(T - 15.4)$

so we have

$16.1 T = 277.87$

$T = 17.26 ^oC$

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F = $F_{1}+ F_{2}$

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= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

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F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

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= - 5.04 x $10^{-17}$ J

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Answer:

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