A 50,0 g silver spoon at 20.0°C is placed in a cup of coffee at

Physics

1 answer:

My name is Ann [436]2 years ago

8 0

The heat capacity of the silver spoon at the given temperature difference is 817.65 J.

<h3>Heat capacity of the silver spoon</h3>

The heat capacity of theb silver spoon is the quantity of heat absorbed by the silver spoon. The heat capacity of the silver spoon at the given temperature difference is calculated as follows;

Q = mcΔθ

where;

m is mass of the spoon
c is specific heat capacity of silver = 0.237 J/g⁰C
Δθ is change in temperature

Q = 50 x 0.237 x (89 - 20)

Q = 817.65 J

Thus, the heat capacity of the silver spoon at the given temperature difference is 817.65 J.

Learn more about heat capacity here: brainly.com/question/16559442

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Explanation:

There seems to be a typo in the problem statement. It says the spring stretches to a shorter length after more mass is added. Please check the problem statement. I'm going to do the calculations assuming that the first length should be 11.80 cm and the second length should be 12.05 cm.

Hooke's law states that the force needed to compress or extend a linear spring is:

F = kΔx, where k is the stiffness and Δx is the displacement.

When a 20g object is placed in the pan, the spring stretches to a length of 11.80 cm. The force of the spring is counteracting the weight of both the pan and the object. Therefore:

(m + 0.020) g = k (0.1180 - 0.100)

And when another 30g object is placed in the pan, the spring stretches to a length of 12.05 cm.

(m + 0.020 + 0.030) g = k (0.1205 - 0.100)

We now have two equations and two variables. If we divide the second equation by the first equation:

(m + 0.050) / (m + 0.020) = (0.1205 - 0.100) / (0.1180 - 0.100)

(m + 0.050) / (m + 0.020) = 0.02050 / 0.0180

0.0180 (m + 0.050) = 0.02050 (m + 0.020)

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