A 39 g slug of silver (specific heat capacity = J/Cg) isheated to 92.1 C and pla

Question

A 39 g slug of silver (specific heat capacity = J/Cg) isheated to 92.1 C and placed in water cooled to 20.0 C (spacificheat4.18 J/cg) if the final temp of the water and silver slug is25.7 C calculat the colume of water the slug was throuninto. ( assume a density of 0.98 g/ml for the water) A 39 g slug of silver (specific heat capacity = J/Cg) isheated to 92.1 C and placed in water cooled to 20.0 C (spacificheat4.18 J/cg) if the final temp of the water and silver slug is25.7 C calculat the colume of water the slug was throuninto. ( assume a density of 0.98 g/ml for the water)

Explanation / Answer

mass of silver = 39 g Specific heat of silver = 0.233 J/ g.C Temperature difference = 92.1 - 25.7 = 66.4 C Heat given out = Mass * specific heat * temperaturedifference                       = 39 g * 0.233 J/ g. C * 66.4 C                        =603.37 J mass of water = m g Specific heat = 4.18 J/ g C Temperature difference = 25.7 - 20 = 5.7 C Heat absorbed = m g * 4.18 J/ g. C * 5.7 C Since heat absorbed by water = Heat given out by thesilver, m g * 4.18 J/ g. C * 5.7 C = 603.37 J So mass of water = 25.32 g density of water = 0.98 g/mL So volume = Mass / density                 = 25.32 g / 0.98 g/ mL                  =25.83 mL

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