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?
- WhomeLv 71 month agoFavorite Answer
½mv² = mcΔT
ΔT = v²/2c = 25² / (2(387)) = 0.80749... ≈ 0.81°C
- Old Science GuyLv 71 month ago
KE = thermal Q
1/2 m v^2 = m c delta(T)
1/2 (25^2) = 387 delta(T)
delta(T) = 0.8075 C°
or 0.81 C° to 2 sig figs
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