Fifty milliliters of 100% H 2 SO 4 at 25 ° C and 84.2mL of liquid water at 15 °
ID: 501299 • Letter: F
Question
Fifty milliliters of 100% H 2 SO 4 at 25 ° C and 84.2mL of liquid water at 15 ° C are mixed. The heat capacity of the product solution is 2.43J/(g ° C).
(a) Estimate the maximum temperature attainable by the product solution and state the conditions under which this temperature would be attained, using heat of mixing data from Table B.11.
(b) Give several reasons why the temperature calculated in Part (a) could not be attained in practice.
(c) Estimate how much heat would have to be transferred from the mixing vessel to keep the temperature of the product solution at 25 ° C. (You should be able to solve this problem quickly by looking back at the enthalpy table of Part (a).)
Explanation / Answer
Given
H2SO4
V1 = 50 ml
T1 = 25 C
density of H2SO4 = 1.84 g/ml
mass m1 = 50 ml * 1.84 g/ml = 92 g
Cp1 = 1.38 J/g.C
water
V2 = 84.2 ml
T2 = 15 C
density = 1g/ml
Mass = 84.2 ml * 1 g/ml = 84.2 g
Cp2 = 4.184 J/g.C =
mixture
taking reference temperature T0 = 273 K = 0 C
total mass m = m1 + m2 = 92 + 84.2 g = 176.2 g
Cp = 2.43 J/g.C
qmix = q1 + q2
m Cp (Tmix - T0) = m1 * Cp1 ( T1 - T0) + m2 * Cp2 *(T2-T0)
176.2 g * 2.43 J/g.C * (Tmix - 0) = 92 g * 1.38 J/g.C ( 25 - 0) C + 84.2 g* 4.184 J/g.C * ( 15 - 0)
176.2 g * 2.43 J/g.C * (Tmix - 0) = 8458.4 J
Tmix - 0 = 19.755 C
Tmix = 19.75 C Answer
reasons are heat will be released to the environment in the process
Heat required to keep it at 25 C = m * Cp * ( 25 - Tmix) = 176.2 g * 2.43 J/g.C * ( 25 - 19.75) = 2247.87 J
Answer 2247.87 J or 2.25 Kj
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