A 1189.4 g aluminium rod at 108.4 °C is placed in 1342.4 ml of water at 16 °C.The specific heat capacities of aluminium and water are 880 J kg-1 K-1 and 4186 J kg-1 K-1. Calculate the final temperature (in °C) of the aluminium block.

Question

A 1189.4 g aluminium rod at 108.4 °C is placed in 1342.4 ml of water at 16 °C.The specific heat capacities of aluminium and water are 880 J kg-1 K-1 and 4186 J kg-1 K-1. Calculate the final temperature (in °C) of the aluminium block.

BlackTom AI · Accepted Answer

Use conservation of energy: heat lost by the hot aluminium equals heat gained by the cooler water (no heat loss to surroundings).

1) Relevant knowledge and conversions
- Heat exchanged: Q = m c ΔT.
- Mass conversions: 1189.4 g = 1.1894 kg (aluminium). 1342.4 ml water ≈ 1342.4 g = 1.3424 kg (density of water = 1.00 g ml−1).
- Specific heats: c_Al = 880 J kg−1 K−1, c_water = 4186 J kg−1 K−1.
- Initial temperatures: T_Al,i = 108.4 °C, T_w,i = 16.0 °C. Final common temperature = T_f.

2) Set up energy balance and rearrange
Heat lost by Al = heat gained by water:
m_Al c_Al (T_Al,i − T_f) = m_w c_w (T_f − T_w,i)

Rearrange to solve for T_f:
m_Al c_Al T_Al,i + m_w c_w T_w,i = T_f (m_Al c_Al + m_w c_w)
so
T_f = (m_Al c_Al T_Al,i + m_w c_w T_w,i) / (m_Al c_Al + m_w c_w)

3) Insert numbers and show intermediate arithmetic
m_Al c_Al = 1.1894 kg × 880 J kg−1 K−1 = 1046.672 J K−1
m_w c_w  = 1.3424 kg × 4186 J kg−1 K−1 = 5619.2864 J K−1

Numerator terms:
m_Al c_Al T_Al,i = 1046.672 × 108.4 = 113459.2448 J
m_w c_w T_w,i   = 5619.2864 × 16.0  = 89908.5824 J

Sum (numerator) = 113459.2448 + 89908.5824 = 203367.8272 J

Denominator = 1046.672 + 5619.2864 = 6665.9584 J K−1

T_f = 203367.8272 / 6665.9584 = 30.508... °C

4) Why this is correct
The final temperature is derived directly from energy conservation with correct unit conversions and use of specific heats. The arithmetic above yields T_f ≈ 30.51 °C.

5) Common incorrect approaches (why they’re wrong)
- Using masses in grams without converting to kg when multiplying by specific heat in J kg−1 K−1 (unit mismatch).
- Forgetting to convert ml of water to kg (assuming different density).
- Setting up the energy equation with the wrong sign (adding rather than equating lost and gained heat).
Any such mistakes lead to different (incorrect) numeric results.

Conclusion: The final temperature of the aluminium block is 30.51 °C, which matches the provided answer.

A 1189.4 g aluminium rod at 108.4 °C is placed in 1342.4 ml of water at 16 °C.The specific heat capacities of aluminium and water are 880 J kg-1 K-1 and 4186 J kg-1 K-1. Calculate the final temperature (in °C) of the aluminium block.数值题

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