Three streams of products (all at 80 °C) mix together in a mixer. The first stream enters at the rate of 10 kg/s and has a specific heat of 4,000 J/kg K. The second stream is pure water, and it enters at the rate of 3 kg/s. The third stream has a specific heat of 3,600 J/kg K. The resulting mixture (product A) enters an indirect contact cooler that uses water at 5 kg/s. The cooling water enters at 5°C and leaves at 25°C. Another product at 40°C (specific heat = 3,000 J/kg K) flowing at the rate of 4 kg/s also enters this cooler and mixes with product A. If the final product (product B) leaves the cooler at the rate of 21 kg/s, determine its temperature. Take into account the fact that the specific heat of water varies with temperature. cp(1)= specific heat of water at 80 °C = 4194 J/kg K cp(3) = specific heat of water at 5 °C = 4206 J/kg K cp(4) = specific heat of water at 25 °C = 4178 J/kg

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
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Please include your diagram, mass balances, and full work. Thank you.

Three streams of products (all at 80 °C) mix together in a mixer. The first stream enters at the
rate of 10 kg/s and has a specific heat of 4,000 J/kg K. The second stream is pure water, and it
enters at the rate of 3 kg/s. The third stream has a specific heat of 3,600 J/kg K. The resulting
mixture (product A) enters an indirect contact cooler that uses water at 5 kg/s. The cooling water
enters at 5°C and leaves at 25°C. Another product at 40°C (specific heat = 3,000 J/kg K) flowing
at the rate of 4 kg/s also enters this cooler and mixes with product A. If the final product (product
B) leaves the cooler at the rate of 21 kg/s, determine its temperature. Take into account the fact
that the specific heat of water varies with temperature.
cp(1)= specific heat of water at 80 °C = 4194 J/kg K
cp(3) = specific heat of water at 5 °C = 4206 J/kg K
cp(4) = specific heat of water at 25 °C = 4178 J/kg
Transcribed Image Text:Three streams of products (all at 80 °C) mix together in a mixer. The first stream enters at the rate of 10 kg/s and has a specific heat of 4,000 J/kg K. The second stream is pure water, and it enters at the rate of 3 kg/s. The third stream has a specific heat of 3,600 J/kg K. The resulting mixture (product A) enters an indirect contact cooler that uses water at 5 kg/s. The cooling water enters at 5°C and leaves at 25°C. Another product at 40°C (specific heat = 3,000 J/kg K) flowing at the rate of 4 kg/s also enters this cooler and mixes with product A. If the final product (product B) leaves the cooler at the rate of 21 kg/s, determine its temperature. Take into account the fact that the specific heat of water varies with temperature. cp(1)= specific heat of water at 80 °C = 4194 J/kg K cp(3) = specific heat of water at 5 °C = 4206 J/kg K cp(4) = specific heat of water at 25 °C = 4178 J/kg
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