Question

A student drops two metallic objects into a 120-g steel container holding 150 g of water at 25°C. One object is a 244-g cube of copper that is initially at 90°C, and the other is a chunk of aluminum that is initially at 5.0°C. To the surprise of the student, the water reaches a final temperature of 25°C, precisely where it started. What is the mass of the aluminum chunk? Answer is in kg.

Answer 1

Answer:Certainly, let's solve the problem.

We can use the principle of conservation of energy to solve this problem. The heat lost by the hot object will be equal to the heat gained by the cold object and the water.

The heat lost by the copper cube can be calculated as:

Q1 = m1 * c1 * (T1 - Tfinal)

where

m1 = 244 g = 0.244 kg (mass of copper cube)

c1 = 0.385 J/g°C (specific heat of copper)

T1 = 90°C (initial temperature of copper cube)

Tfinal = 25°C (final temperature of the system)

Substituting the values, we get:

Q1 = 0.244 * 0.385 * (90 - 25) = 21.38 J

Similarly, the heat gained by the aluminum chunk can be calculated as:

Q2 = m2 * c2 * (Tfinal - T2)

where

m2 = ? (mass of aluminum chunk)

c2 = 0.902 J/g°C (specific heat of aluminum)

T2 = 5.0°C (initial temperature of aluminum chunk)

Substituting the values, we get:

Q2 = m2 * 0.902 * (25 - 5.0) = 18.144 m2 J

Now, since the water temperature doesn't change, we can assume that the heat gained by the water is equal to the heat lost by the hot objects. Therefore, we can write:

Q1 + Q2 = mwater * cwater * (Tfinal - Tinitial)

where

mwater = 150 g = 0.15 kg (mass of water)

cwater = 4.184 J/g°C (specific heat of water)

Tinitial = 25°C (initial temperature of water)

Substituting the values, we get:

21.38 J + 18.144 m2 J = 0.15 * 4.184 * (25 - 25)

Simplifying the equation, we get:

m2 = 0.266 kg

Therefore, the mass of the aluminum chunk is 0.266 kg or 266 grams.

Step-by-step explanation:

Answer 2

Therefore, the mass of the aluminum chunk is 15 g or 0.015 kg.

Step-by-step explanation:

We have a container with 120 g of steel and 150 g of water at 25°C.

Two metal objects are dropped into the container:

A 244 g cube of copper initially at 90°C

An unknown mass of aluminum initially at 5.0°C

The surprise result is that the final water temperature is still 25°C.

Since the final water temperature is the same, the heat lost by the hot copper must equal the heat gained by the cold aluminum and the container.

The heat lost by the copper can be calculated as:

Qcopper = copper * copper * ΔTcopper

= 0.244 kg * 385 J/(kg•K) * (90°C - 25°C) = 29.5 kJ

The heat gained by the aluminum and container can be calculated as:

Qtotal = maluminum * cpaluminum * ΔTaluminum + mcontainer * cpcontainer * ΔTcontainer

Setting this equal to 29.5 kJ and plugging in the given values:

29.5 kJ = maluminum * 900 J/(kg•K) * (25°C - 5°C) + 0.120 kg * 450 J/(kg•K) * (25°C - 25°C)

Solving for aluminum :

aluminum = 0.015 kg

aluminum = 15 g