One way to cool a cup of coffee is to plunge an ice-cold piece of aluminum into

Question

One way to cool a cup of coffee is to plunge an ice-cold piece of aluminum into it. Suppose a 20.0 g piece of aluminum is stored in the refrigerator at 0 C and then put into a cup of coffee. The coffee's temperature drops from 90.0 C to 75.0 C. (A) How much energy in joules did the coffee transfer to the piece of aluminum? **I calculated this already, it's 1353 J (1.353 kJ)** (B) Using the same information, calculate the mass (g) of coffee in the cup (Specific Heat Coffee = Specific Heat Water = 4.18 J/g*C).

Explanation / Answer

As per your calculation,

Heat transferred by coffee = 1353 J

Ans-B)

For a coffee,

Q = – 1353 J ……. (As heat transferred from system to surrounding.)

Specific heat capacity for coffee = Sp. Heat capacity of water = Cp = 4.18 J/g oC

Initial temperature Ti = 90 oC

Final temperature Tf = 75 oC

Change in temperature (T) = Tf – Ti = 75 – 90 = –15 oC

Mass of coffee = m = ? g

Q, Cp, m and T are related by the relation,

Q = m x Cp x T

– 1353 = m x 4.18 x (–15)

1353 = m x 4.18 x 15

m = 1353 /(4.18 x 15)

m = 21.58 g

Mass of coffee is 21.58 g

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