Answer:The quantity of warm released once 50 g the water cooled native 20°C to 10°C will be equal to - 2093 J.Explanation:Given data:Mass that water = 50 gInitial temperature= T1 = 20°CFinal temperature= T2 = 10°CSpecific warm of water= c = 4.186 J/g. °CAmount of warm released = Q= ?Solution:Formula:Q = m. C. ΔTΔT = T2 - T1ΔT = 10°C - 20°CΔT = -10°CNow we will placed the worths in formula.Q = m. C. ΔTQ = 50 g . 4.186 J/g. °C . -10°CQ = - 2093 JThe quantity of warmth released once 50 g the water cooled native 20°C come 10°C will certainly be equal to - 2093 J.

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