When 50 g of water at 90 °C is added to a calorimeter containing 50 g of water at 15 °C, the temperature increases to 45 °C - VCE - SSCE Chemistry - Question 14 - 2012 - Paper 1

The energy gained by the cold water is given by the same formula as above, but for the cold side:

$Q_{cold} = m_{cold} imes c_{water} imes (T_{final} - T_{initial, cold})$

Where:

• $m_{cold} = 50 ext{g}$
• $c_{water} = 4.18 ext{J/g°C}$
• $T_{initial, cold} = 15 °C$
• $T_{final} = 45 °C$

Calculating: $Q_{cold} = 50 imes 4.18 imes (45 - 15) = 50 imes 4.18 imes 30 = 6270 ext{J}$

Thus, energy gained by the cold water is equal to 6270 J. This energy corresponds directly to the energy lost by the hot water, demonstrating that energy is conserved.

Hence, the answer is B.

Based on the principles of conservation of energy, the total energy absorbed by the calorimeter equals the sum of energy lost by the hot water and energy gained by the cold water:

$Q_{total} = Q_{cold} + Q_{hot}$

From previous calculations:

• $Q_{hot} = 9390 ext{J}$
• $Q_{cold} = 6270 ext{J}$

If we sum them: $Q_{total} = 9390 + 6270 = 15660 ext{J}$

This confirms that the total is equivalent to the energy involved in the exchange. However, instead, the question focuses on the direct correlation:

The answer is C, emphasizing that in reality, the total energy exchanged relates back to the first two options.

The difference between the energy lost by the hot water and gained by the cold water can be described mathematically as:

$ext{Difference} = Q_{hot} - Q_{cold}$

Using previous calculations:

• $Q_{hot} = 9390 ext{J}$
• $Q_{cold} = 6270 ext{J}$

Calculating the difference gives: $ext{Difference} = 9390 - 6270 = 3120 ext{J}$

This demonstrates that there is indeed a residual energy which could be attributed to external factors in a real-world scenario. Hence, while the question is misattributed in part D in typical phrasing, it's transformative to see the response under its significant interpretations.

This continued clarity ensures the answer is D.