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heat is lost to the styrofoam calorimeter. assuming a 6.22 c temperature change for the reaction of hcl (aq) with naoh (aq), calculate the heat loss to the inner 2.35g styrofoam cup. the specific heat of styrofoam is 1.34 j/g*c

Sagot :

Given in question that;

change in temperature; ∆T = 6.22°c

weight of substance; W = 2.35 g

specific heat of styrofoam; Cp = 1.34 j/g°c

To find heat loss; Q = w x Cp x ∆T

{where, change in temperature; ∆T}

{weight of substance; W}

{specific heat of 2.35gmaterial; Cp}

Thus, Q = 2.35 g × 6.22°c × 1.34 j/g°c

          Q = 19.58 j

so, 19.6 j heat is lost

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