A large pot (10 kg) of beef stew is left out on the kitchen counter to cool. The rate of cooling, in kJ/min, equals 1.955 (T-25), where T is the temperature of the stew in degrees Celsius. The heat capacity of the stew is 4 kJ/kg-C. If the stew initially is at 90 C, how long does it take to cool to 40 C? Round off your answer to the nearest minute.

Question

Elisa Mcmillan

Chemistry

16 December, 02:38

A large pot (10 kg) of beef stew is left out on the kitchen counter to cool. The rate of cooling, in kJ/min, equals 1.955 (T-25), where T is the temperature of the stew in degrees Celsius. The heat capacity of the stew is 4 kJ/kg-C. If the stew initially is at 90 C, how long does it take to cool to 40 C? Round off your answer to the nearest minute.

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Answers (1)

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Gideon · Answer 1

16 minutes

Explanation:

First, we need to calculate the amount of heat needed to cool the beef stew:

Q = mcΔT

Where m is the mass, c is the heat capacity and ΔT is the variation of the temperature.

Q = 10x4x (40 - 90)

Q = - 2000 kJ

So, the beef stew needs to lost 2000 kJ to cool.

With the initial temperature at 90ºC, the rate of cooling (r) will be:

r = 1.955x (90 - 25)

r = 127.075 kJ/min

So, to lose 2000 kJ, will be necessary:

t = Q/r

t = 2000/127.075

t = 16 minutes