A 2.50-kg block of hot iron (cFe = 0.45 J/g K; Tin = 300. oC) is dropped in cold

ID: 949628 • Letter: A

Question

A 2.50-kg block of hot iron (cFe = 0.45 J/g K; Tin = 300. oC) is dropped in cold water (Tin = 20 oC) to cool quickly.

1) how much heat needs to be absorbed to cool the iron block to 25 oC?

A) 309 J B) 309 kJ C) 618 J D) 618 kJ E) None of the above

2) How much cold water will be needed?

A) 14.8 g B) 14.8 kg C) 14.8 m3 D) 14,800 L E) none of the above

3) What will happen if 10.0 L of cold water are used?

A) The water will start boiling

B) Only part of the metal will be cooled down

C) The metal will be cooled down but only to a higher T than 25

D) The water will end up at a higher temperature than the metal

E) None of the above

Mass of Iron (m) = 2.5 kg = 2500 g

Heat capacity of iron (Cp) = 0.45 J / g K

Initial Temperature of Iron (T1)= 300 degree C

1) Final Temperature of Iron (T2) = 25 degree C

Heat released by Iron = m Cp (T1 - T2)

=> Heat = 2500 x 0.45 x (300 - 25) = 309375 J = 309.375 kJ

B) 309 kJ

Initial Temperature of cold water (T1) = 20 degree C

Final Temperature of cold water (T2) = 25 degree C

Cp for cold water = 4.184 J / g K

Heat absorbed = 309375 J

=> 309375 =m Cp (T2 - T1)

=> 309375 = m x 4.184 x 5

=> m = 14788.5 g = 14.8 kg

B) 14.8 kg

10 L water = 10 kg water, which is less than the required 14.8 kg.

Therefore,

C) The metal will be cooled down but only to a higher T than 25

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