Total change in entropy, what am I missing?

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To estimate the total change in entropy of the universe when a 20.0-kg box slides to rest, the kinetic energy lost by the box is converted into heat, which is absorbed by the table and room at a constant temperature of 293 K. The kinetic energy of the box is calculated using the formula KE = 0.5 * m * v^2, resulting in a value of 10 J. Since the box does not lose heat, its entropy change is zero, while the table and room experience an increase in entropy due to the absorbed heat. The total change in entropy can be calculated using the formula ΔS = Q/T, leading to a positive value for the universe's entropy. Understanding these principles clarifies the process of estimating entropy changes in thermodynamic systems.
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A 20.0 -kg box having an initial speed of 1.0 m/s slides along a rough table and comes to rest.
Estimate the total change in entropy of the universe. Assume all objects are at room temperature (293 K).

I don't understand how to answer this problem. It feels like I wasn't given enough information to solve for Delta S.
 
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asdf188 said:
A 20.0 -kg box having an initial speed of 1.0 m/s slides along a rough table and comes to rest.
Estimate the total change in entropy of the universe. Assume all objects are at room temperature (293 K).

I don't understand how to answer this problem. It feels like I wasn't given enough information to solve for Delta S.
Assume that the kinetic energy of the box is dissipated as heat and that this heat is absorbed by the table and room at a constant temperature (ie the temperature of the table and room does not change).

Hint: Since the box does not lose any heat, there is 0 change in the entropy of the box. However, the table gains heat.

AM
 
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