Question

An inexperienced clinician was measuring the cardiac output of a healthy human by thermodilution technique. A 2.0 mL of cold saline solution of volume \( V_i \) at \( 7^\circ C \) was injected at the entrance of the right atrium. The change in blood temperature \( \int_0^{t_1} \Delta T_b \, dt \) at the pulmonary artery was measured to be \(-20\) Kelvin-second. The cardiac output \( F \) was calculated using the following formula \[ F = \frac{Q}{\rho_b c_b \int_0^{t_1} \Delta T_b \, dt}, \] where \( Q \) is the heat content of injectate in Joules, given by \( V_i \Delta T_i \rho_i c_i \) and \( \Delta T_i \) is the temperature difference between the injectate and blood. It was assumed that the density of blood (\( \rho_b \) in kg/m\(^3\)) and the specific heat capacity of blood (\( c_b \) in J/(kg.K)) were respectively equal to that of the injectate \( \rho_i \) and \( c_i \).
The clinician realized that there was an error in the measurement of \( F \). Which of the following is TRUE?

• A. Cardiac output is too low because the cold saline volume was too small
• B. Cardiac output is too low because \( \int_0^{t_1} \Delta T_b \, dt \) is too large
• C. Cardiac output is too high because the cold saline volume was too large
• D. Cardiac output is too high because \( \int_0^{t_1} \Delta T_b \, dt \) is too small

Hint:

The cardiac output is inversely proportional to the volume of the injected cold saline, so a smaller volume leads to a smaller measured output.

Answer 1

The Correct Option is A, B

In the formula for cardiac output: \[ F = \frac{Q}{\rho_b c_b \int_0^{t_1} \Delta T_b \, dt}, \] the output \( F \) depends on the value of the temperature change \( \Delta T_b \) and the volume of the injected saline (via \( Q \)). Analysis of the error: - If the cold saline volume is too small (which is the case here with 2.0 mL), the total heat content \( Q \) will be too small. Since \( Q \) is in the numerator of the formula, this leads to a smaller cardiac output. - If the temperature change \( \Delta T_b \) is too large, it would increase the denominator, thus leading to a smaller cardiac output, but this is not the case based on the measurements given. Thus, the correct answer is Option (A): the cardiac output is too low because the cold saline volume was too small.