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puteri [66]
3 years ago
11

A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i

ron cube was 20.0 ˚C. The cube of iron raised the temperature of the water to 45.0 ˚C. If the iron cube has a mass of 133 g, what is the specific heat capacity of iron?
Physics
1 answer:
Mariana [72]3 years ago
4 0

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

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In an alpha decay, an atom emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons: this means that during this kind of decay, the original atom loses 2 protons and 2 neutrons from its nucleus.

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A resistor has four colored stripes in the following order: orange, orange, brown and silver. What is the resistance of the resi
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Answer:

Resistance =330 Ω

Tolerance = 33 Ω

Explanation:

see attached resistor color code table

The first stripe is orange, which means the leftmost digit is a 3.

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The resistance is:

orange-orange-brown=  330 Ω

The tolerance is:

The fourth color band indicates the resistor's tolerance.  Tolerance is the percentage of error in the resistor's resistance.

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A 330 Ω resistor has a silver tolerance band.  

<em>Tolerance = value of resistor x value of tolerance band </em>

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330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.

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A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g
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For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

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X = 1.4m

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Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

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1.4 - 4.905t² = 0

4.905t² = 1.4

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t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

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