Respuesta :

superman1987
when the heat lost by water = heat gained by ethanol

∴(  M* C * ΔT )w  = (M*C*ΔT ) eth

when Mw mass of water = 40 g 

C specific heat of water = 4.18

ΔT = (70- Tf) 

and M(eth) mass (ethanol) = 40 g 

C specific heat of ethanol = 2.44 

ΔT = (Tf - 10 ) 

by substitution:

40* 4.18 * (70 - Tf) = 40 * 2.44 * (Tf-10) 

∴Tf = 47.9 °C 

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