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What is the specific heat of a metal if a ten gram sample at 40.2°C is dropped into 100 grams of water at 25.4°C and the temperature rises to 28.6°C? (Sp. heat of H_2O = 4.18 J/g•°C)

a. 2.11 J/g °C
b. 0.085 J/g °C
c. 206 J/g °C
d. 1.92 J/g °C
e. 20.9 J/g °C

1 Answer

5 votes

Final answer:

The specific heat capacity of the metal (c_m) is calculated using the principle of conservation of energy and the formula q = mcΔT. After setting up the equation based on the given data and solving it, we find that c_m equals 0.137 J/g°C, which does not match any of the given options.

Step-by-step explanation:

We can solve this problem by applying the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water. We will use the formula q = mcΔT, where q is the amount of heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the temperature change.

Let c_m represent the specific heat capacity of the metal. For the metal, the heat lost is q_m = m_m × c_m × (ΔT_m), and for the water, the heat gained is q_w = m_w × c_w × (ΔT_w). Since the heat lost by metal is gained by the water, q_m = -q_w.

The mass of the water (m_w) is 100 grams and its specific heat capacity (c_w) is 4.18 J/g°C. The change in water's temperature (ΔT_w) is 3.2°C (from 25.4°C to 28.6°C).

The mass of the metal (m_m) is 10 grams, and its temperature change (ΔT_m) is -40.2°C (from 40.2°C to the final temperature of 28.6°C).

Now we can set up the equation:
m_m × c_m × (-ΔT_m) = m_w × c_w × ΔT_w,
10g × c_m × (-11.6°C) = 100g × 4.18 J/g°C × 3.2°C,
c_m = [100g × 4.18 J/g°C × 3.2°C] / [10g × -11.6°C].

After solving, we get c_m = 0.137 J/g°C, which is not listed in the options provided and suggests an error in the question or the options given.

answered
User Asiya
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