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3 years ago

7

A 0.5-kilogram piece of aluminum increases its temperature 7°C when heat energy (Q) is added. Set up the equation with the corre

ct values.

1 answer:

mars1129 [50]3 years ago

6 0

Answer:

Heat energy (Q) added is 3150 (J) .

Explanation:

Given:

Mass of the aluminum = 0.5 kg

Change in temperature = 7°C

According to the question :

Heat energy (Q) is added up.

We have to find the equation with correct values.

We know that,Q = mCΔT m is the mass,C is the specific heat and T is the change in temperature.

Specific heat of aluminum = 900 (J/kg °C)

Plugging the values,we have :

⇒ $Q=mC\Delta T$

⇒ $Q = 0.5\times 900\times 7$

⇒ $Q= 3150$ Joules (J)

So the heat energy added (Q) = 3150 Joules.

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Find the temperature of a mixture obtained by mixing 32 grams of water at a temperature of 22 degrees Celsius with 88 grams of w

fomenos

Answer:

60.87 °C

Explanation:

Applying,

Heat lost = heat gain

cm'(t₂-t₃) = cm(t₃-t₁).............. Equation 1

Equation 1 can futher be simplified to

m'(t₂-t₃) = m(t₃-t₁)................Equation 2

Where m' = mass of the hot water, m = mass of the cold water, t₁ = initial temperature of the cold water, t₂ = initial temperature of the hot water, t₃ = temperature of the mixture.

From the question,

Given: m' = 88 g, m = 32 g, t₁ = 22°C, t₂ = 75°C

Substitute these values into equation 2

88(75-t₃) = 32(t₃-22)

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120t₃ = 7304

t₃ = 60.87 °C

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A 10 kg block rests on a 30o inclined plane. The block is attached to a bucket by pulley system depicted below. The mass in the

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Answer:

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b). am = 0.56 $m/s^2$ (block), aM = 0.28 $m/s^2$ (bucket)

Explanation:

a). We got N = mg cos θ,

f = $\mu_s N$

= $\mu_s mg \cos \theta$

If the block is ready to slide,

T = mg sin θ + f

T = mg sin θ + $\mu_s mg \cos \theta$ .....(i)

2T = Mg ..........(ii)

Putting (ii) in (i), we get

$\frac{Mg}{2}=mg \sin \theta + \mu_s mg \sin \theta$

$M=2(m \sin \theta + \mu_s mg \cos \theta)$

$M=2 \times 10 \times (\sin 30^\circ+0.6 \cos 30^\circ)$

M = 20.392 kg

b). $(h-x_m)+(h-x_M)+(h'+x_M)=l$ .............(iii)

Here, l = total string length

Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so

$-\ddot{x}-2\ddot x_M=0$

$\ddot x_M=\frac{\ddot x_m}{2}$

$a_M=\frac{a_m}{2}$ .....................(iv)

We got, N = mg cos θ

$f_K=\mu_K mg \cos \theta$

∴ $T-(mg \sin \theta + f_K) = ma_m$

$T-(mg \sin \theta + \mu_K mg \cos \theta) = ma_m$ ................(v)

Mg - 2T = M $a_M$

$Mg-Ma_M=2T$

$Mg-\frac{Ma_M}{2} = 2T$ (from equation (iv))

$\frac{Mg}{2}-\frac{Ma_M}{4}=T$ .....................(vi)

Putting (vi) in equation (v),

$\frac{Mg}{2}-\frac{Ma_M}{4}-mf \sin \theta-\mu_K mg \cos \theta = ma_m$

$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$

$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$

$a_m= 0.56 \ m/s^2$

Using equation (iv), we get,

$a_M= 0.28 \ m/s^2$

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