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An 80.0 g sample of metal, initially at 96.0 °C, is placed into 150. g of water initially at 26.0 °C in a calorimeter. The final temperature of the water is 28.1 °C. What is the identity of the metal? (The specific heat of water is 4.18 J/g°C.)

A) Fe
B) Sn
C) Ag
D) Cu

User Rajul
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1 Answer

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Final answer:

The specific heat capacity of the metal is approximately 0.17146 J/g°C. The identity of the metal is Silver (Ag).

Step-by-step explanation:

To determine the specific heat and identity of the metal, we can use the principle of heat transfer in a calorimeter. The heat gained by the water is equal to the heat lost by the metal. We can use the formula:

qwater = qmetal

where qwater is the heat gained by the water, qmetal is the heat lost by the metal. The formula for heat is:

q = mcΔT

where q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. By rearranging the equations and substituting the given values, we can calculate the specific heat and identify the metal.

Let's start with the equation for the water:

qwater = mwatercwaterΔTwater

qwater = (150.0 g)(4.18 J/g°C)(28.1 °C - 26.0 °C) = 1254.6 J

Now, let's calculate the heat lost by the metal:

qmetal = mmetalcmetalΔTmetal

qmetal = (80.0 g)(cmetal)(28.1 °C - 96.0 °C)

Setting the heat gained and lost equal to each other:

mwatercwaterΔTwater = mmetalcmetalΔTmetal

(150.0 g)(4.18 J/g°C)(28.1 °C - 26.0 °C) = (80.0 g)(cmetal)(28.1 °C - 96.0 °C)

1254.6 J = -7325.6 g°C(cmetal)

Simplifying:

cmetal = -1254.6 J / (-7325.6 g°C)

cmetal ≈ 0.17146 J/g°C

The specific heat capacity of the metal is approximately 0.17146 J/g°C. To determine the identity of the metal, we can compare this value to the specific heat capacities of known metals. The metal with a specific heat capacity closest to 0.17146 J/g°C is Silver (Ag).

User Nizar
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