Before going in for an annual physical, a 70.0-kg person whosebody temperature i

Question

Before going in for an annual physical, a 70.0-kg person whosebody temperature is 37.0 degrees C consumes an entire0.355-liters can of a soft drink (which is mostly water) at 12.0degrees C. What will be the person's body temperature Tfinal afterequilibrium is attained? Ignore any heating by the person'smetabolism. The specific heat capacity of a human body is 3480 J/kgx K. Before going in for an annual physical, a 70.0-kg person whosebody temperature is 37.0 degrees C consumes an entire0.355-liters can of a soft drink (which is mostly water) at 12.0degrees C. What will be the person's body temperature Tfinal afterequilibrium is attained? Ignore any heating by the person'smetabolism. The specific heat capacity of a human body is 3480 J/kgx K.

Explanation / Answer

Initial heat content of the person's body is : Q = M*C*T M = 70.0 kg C = 3480 J/kg-K T = 37.0 + 273 K = 310 K So, Q = 75,516,000 J = 7.5516*107 J Heat content of the soft drink : Q' = M'*C'*T' M' = 0.355 liters of water = 0.355 kg of water C' = 4184 J/kg-K T' = 12.0 + 273 K = 285 K So, Q' = 423,316.2 J = 4.233162*105 J So, total heat content = Q + Q' = Q" = 7.5939*107 J, where Q" is the final heat content at equilibrium of the twosystems combined. So, if the final temperature at equilibrium is T", Q" = M*C*T" + M'*C'*T"      = [72.0*3480 + 0.355*4184] * T" =7.5939*107 J So, T" = final equilibrium temperature = 301.29 K = 301.29 - 273 K = 28.29 degrees C

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