29g of copper pellets are removed from a 300 Celsius oven and immediately dropped into 100 mL of water at 22 Celcius in an insulated cup. What will the new water temperature be?

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29g of copper pellets are removed from a 300 Celsius oven and immediately dropped into 100 mL of water at 22 Celcius in an insulated cup. What will the new water temperature be?

Physics
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Q=m*c*delta(T) simple
That's not quite right. Actually it's: \[\Delta Q = m c \Delta T\] So you know, cause of the conservation of energy, that the quantity of heat remains the same. So: \[c_1 * m_1 * (T_1 - T_*) = c_2 * m_2 * (T_* - T_2)\] when plugging in 0.39 for copper and 4.19 i get 29°C for the resulting temperature...

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