Question

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.

Answer 1

Answer:

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