# what was the increase in temperature?

A copper sphere was moving at 25 m/s when it hit another object. This caused all of the KE to be converted into thermal energy for the copper sphere. If the specific heat capacity of copper is 387 J/(kg ⋅ C°), what was the increase in temperature?

0.23 C°

0.81 C°

1.3 C°

2.1 C°

### 2 Answers

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- WhomeLv 71 month agoFavorite Answer
½mv² = mcΔT

ΔT = v²/2c = 25² / (2(387)) = 0.80749... ≈ 0.81°C

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- Old Science GuyLv 71 month ago
KE = thermal Q

1/2 m v^2 = m c delta(T)

m's cancel

1/2 (25^2) = 387 delta(T)

delta(T) = 0.8075 C°

or 0.81 C° to 2 sig figs

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