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Answer :
Final answer:
The specific heat capacity of aluminum in this example is 0.394 J/g °C. This was determined by calculating the heat gained by water and considering that it equals the heat lost by the aluminum, and subsequently solving for the specific heat capacity of aluminum.
Explanation:
To find the specific heat capacity (C) of aluminum, we must consider the amount of heat transferred from the aluminum to the water (expressed as q). The heat gained by water is calculated using the equation q = m * C * ΔT, where m is the mass, C is the specific heat, and ΔT is the change in temperature. According to the question, water's mass (m) is 50.0g, its specific heat (C) is 4.184 J/g °C, and the difference in its temperature (ΔT) is 2.00 °C. So, the heat gained by water is q = 50.0g * 4.184 J/g * °C * 2.00 °C = 418.4 J.
The heat lost by aluminum is equal to the heat gained by water. Therefore, using the equation q = m * C * ΔT and plug in the values of q (418.4 J), m (28.4 g), and ΔT (39.4 °C), to solve for C, the specific heat of aluminum, we can rearrange the formula to C = q / (m * ΔT) = 418.4 J / (28.4g * 39.4 °C) = 0.394 J/g °C.
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Answer:
C Al = 0.8975 J/g.K
Explanation:
- Q = mCΔT
∴ m Al = 28.4 g
∴ T Al = 39.4°C = 312.4 K
∴ m H2O = 50.0 g
∴ T1 H2O = 21°C = 294 K
∴ T2 H2O = 23°C = 296 K
∴ C H2O = 4,18 J/g.K
⇒ C Al = ?
in a calorimeter:
∴ Al give heat: Q Al < 0
∴ H2O revceives heat: Q H2O > 0
⇒ - Q Al = Q H2O
⇒ - (28.4 g)*(C Al)*(296 K - 312.4 K) = (50.0 g)*(4.18 J/g.K)*(296 K - 294 K)
⇒ - (- 465.76 g.K)*(C Al) = 418 J
⇒ C Al = (418 J) / (465.76 g.K)
⇒ C Al = 0.8975 J/g.K = 897.5 J/Kg.K
