Problem: Your stomach is a chemical reactor. When you consumea fast food $0.99 h

Question

Problem: Your stomach is a chemical reactor. When you consumea fast food $0.99 hamburger in about a minute, it acts like aninstantaneous input of 325 g of food entering the stomach. Inresponse, the stomach starts producting gastric liquids (acids),which are continuously excreted into the stomach at a rate of 12.0mL/min as the hamburger is digested. The fluid also leaves thestomach to the small intestine at a flow rate of 12.0 mL/min, sothe volume of liquid in the stomach stays constant at 1.15 L. Thehamburger digestion rate is constant at 1.33/hr. (a) What kind of ideal reactor would you model your stomachas? (b) What fraction of the hamburger's mass will remainundigested in you stomach one hour after you eat thehamburger? Problem: Your stomach is a chemical reactor. When you consumea fast food $0.99 hamburger in about a minute, it acts like aninstantaneous input of 325 g of food entering the stomach. Inresponse, the stomach starts producting gastric liquids (acids),which are continuously excreted into the stomach at a rate of 12.0mL/min as the hamburger is digested. The fluid also leaves thestomach to the small intestine at a flow rate of 12.0 mL/min, sothe volume of liquid in the stomach stays constant at 1.15 L. Thehamburger digestion rate is constant at 1.33/hr. (a) What kind of ideal reactor would you model your stomachas? (b) What fraction of the hamburger's mass will remainundigested in you stomach one hour after you eat thehamburger?

Explanation / Answer

a) Your stomach acts as a CSTR reactor, which in this case is a first order decay reaction. The CSTR takes gastric juices in, proceeds with a batch reaction, and outputs the mixed fluids to the intestine. b) Using mass balance: Accumulation = Input - Output +/- Reaction and C_initial = 325g/1.15L = 282.6 g/L V(dC/dt) = Q_initial*C_initial - Q_out*C_out - V*k*C (since this is first order decay) ?dC/C from C_i to C_out = -(Q/V + k) ?dt from 0 to t C_out = C_in*e^-((Q/V+k)*t) And the Fraction = C_out/C_in So... C_out = (282.6 g/L)*e^-((0.012L/min*60min/hr)/(1.15L)+1.33hr^-1*1hr) = 39.96 g/L And the Fraction = C_out/C_in = (39.96 g/L) / (282.6 g/L) = 0.141414 = 14/99 undigested

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