Sidney P. answered • 05/19/21
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From Q = m c ΔT, Q = (202.3 g) (4.186 J/g °C) (-10.70 -25.60) = -30,740 J = -30.74 kJ.
Emi M.
asked • 05/18/21What quantity of heat (in kJ) would need to be removed to cool 202.3 g of water from 25.60°C to -10.70°C?
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