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Lubov Fominskaja [6]
2 years ago
10

You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re

cord the Temperature rise from 20C to 40C. What is the specific heat of (AU)Gold?
Physics
1 answer:
melamori03 [73]2 years ago
7 0

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

Quantity of heat applied, Q = 3096 J

Temperature, t2 = 40°C

Temperature, t1 = 20°C

Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

Q = mcdt

Where, C = specific heat capacity of gold

3096 = 1.2kg * C * 20°C

3096 J = 24kg°C * C

C = 3096 J / 24 kg°C

C = 129 J/Kg°C

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Answer:

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Explanation:

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The magnitude of speed of the speed of the boat relative to a stationary shore observer is 1.85 m/s

The distance covered is 229 m

Time = Distance / Speed

\text{Time}=\frac{229}{1.69}\\\Rightarrow \text{Time}=135.50\ s

Distance = Speed of current × Time

Distance = 0.756×135.5 = 102.438 m

The boat will be 102.438 m from the initial position is the boat when it reaches the opposite shore

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<u>Answer</u>

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