A 200 gram copper calorimeter contains 400 grams of water at 25 C. A metal specimen weighing 250 grams is heated to a certain temperature and the plunged into the calorimeter The final temperature of the mixture is 42C. If then specific heat of the metal specimen is 0.09 cal/(g-C), find the initial temperature of the specimen just before it was plunged into the calorimeter

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To find the initial temperature of the metal specimen, we can use the principle of heat transfer.

The heat lost by the metal specimen is equal to the heat gained by the water and the calorimeter.

The heat lost by the metal specimen can be calculated using the formula:

Q1 = m1 * c1 * (T1 - T2)

where:
Q1 = heat lost by the metal specimen
m1 = mass of the metal specimen (250 grams)
c1 = specific heat of the metal specimen (0.09 cal/(g-C))
T1 = initial temperature of the metal specimen (unknown)
T2 = final temperature of the mixture (42 C)

The heat gained by the water and the calorimeter can be calculated using the formula:

Q2 = (m2 + m3) * c2 * (T2 - T3)

where:
Q2 = heat gained by the water and the calorimeter
m2 = mass of the water (400 grams)
m3 = mass of the calorimeter (200 grams)
c2 = specific heat of the water and calorimeter (1 cal/(g-C))
T3 = initial temperature of the water and the calorimeter (25 C)

Since the heat lost by the metal specimen is equal to the heat gained by the water and the calorimeter, we can equate Q1 and Q2:

m1 * c1 * (T1 - T2) = (m2 + m3) * c2 * (T2 - T3)

Plugging in the given values:

250 * 0.09 * (T1 - 42) = (400 + 200) * 1 * (42 - 25)

Simplifying the equation:

22.5 * (T1 - 42) = 600 * (42 - 25)

22.5 * T1 - 945 = 600 * 17

22.5 * T1 - 945 = 10200

22.5 * T1 = 10200 + 945

22.5 * T1 = 11145

T1 = 11145 / 22.5

T1 ≈ 496

Therefore, the initial temperature of the metal specimen just before it was plunged into the calorimeter is approximately 496 degrees Celsius.

To find the initial temperature of the metal specimen, we can use the principle of conservation of energy.

The total heat gained by the water and the calorimeter is equal to the heat lost by the metal specimen.

First, let's calculate the heat gained by the water and the calorimeter:

Qwater = mwater * cwater * ΔTwater

where:
- Qwater is the heat gained by the water and the calorimeter
- mwater is the mass of water (400 grams)
- cwater is the specific heat capacity of water (1 cal/(g-C))
- ΔTwater is the change in temperature of water (final temperature - initial temperature)

Next, let's calculate the heat lost by the metal specimen:

Qmetal = mmetal * cmetal * ΔTmetal

where:
- Qmetal is the heat lost by the metal specimen
- mmetal is the mass of the metal specimen (250 grams)
- cmetal is the specific heat capacity of the metal specimen (0.09 cal/(g-C))
- ΔTmetal is the change in temperature of the metal specimen (final temperature - initial temperature)

Since the heat gained and lost are equal, we have:

Qwater = Qmetal

mwater * cwater * ΔTwater = mmetal * cmetal * ΔTmetal

We can rearrange the equation to solve for the initial temperature of the metal specimen (ΔTmetal):

ΔTmetal = (mwater * cwater * ΔTwater) / (mmetal * cmetal)

Substituting the given values:

ΔTmetal = (400 grams * 1 cal/(g-C) * (42 C - initial temperature)) / (250 grams * 0.09 cal/(g-C))

Now, we can solve for the initial temperature of the metal specimen by rearranging the equation:

initial temperature = 42 C - (ΔTmetal * (250 grams * 0.09 cal/(g-C))) / (400 grams * 1 cal/(g-C))

By calculating the right side of the equation, you can find the initial temperature of the metal specimen before it was plunged into the calorimeter.

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