Question

A hot lump of 27.4 g of aluminum at an initial temperature of 69.5 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.​

Answer 1

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$\red{\bold{{\underline{ Solution \: with\: Explanation \: :}}}}$

$\large \boxed{29.7 \,^{\circ}\text{C}}$

There are two heat transfers involved: the heat lost by the aluminium and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the Al be Component 1 and the H₂O be Component 2.

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$m_{1} =\text{27.4 g; }T_{i} = 69.5 ^{\circ}\text{C; }\\C_{1} = 0.903 \text{ J ^{\circ} C ^{-1} g ^{-1} }$

For the water:

$m_{2} =\text{50.0 g; }T_{i} = 25.0 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J ^{\circ} C ^{-1} g ^{-1} }$

Calculations

(a) The relative temperature changes

$\begin{array}{rcl}\text{Heat lost by Al + heat gained by water} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{27.4 g}\times 0.903 \text{ J ^{\circ} C ^{-1} g ^{-1} } \times\Delta T_{1} + \text{50.0 g} \times 4.184 \text{ J ^{\circ} C ^{-1} g ^{-1} }\Delta \times T_{2} & = & 0\\24.74\Delta T_{1} + 209.2\Delta T_{2} & = & 0\\\end{array}$

(b) Final temperature

$\Delta T_{1} = T_{\text{f}} - 69.5 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 25.0 ^{\circ}\text{C}$

$\begin{array}{rcl}24.74(T_{\text{f}} - 69.5 \, ^{\circ}\text{C}) + 209.2(T_{\text{f}} - 25.0 \, ^{\circ}\text{C}) & = & 0\\24.74T_{\text{f}} - 1719 \, ^{\circ}\text{C} + 209.2T_{\text{f}} -5230 \, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} - 6949\, ^{\circ}\text{C} & = & 0\\233.9T_{\text{f}} & = & 6949 \, ^{\circ}\text{C}\\T_{\text{f}}& = & \mathbf{29.7 \, ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature is \large \boxed{\mathbf{29.7 \,^{\circ}}\textbf{C}} }$

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Check:

$\begin{array}{rcl}27.4 \times 0.903 \times (29.7 - 69.5) + 50.0 \times 4.184 (29.7 - 25.0)& = & 0\\24.74(-39.8) +209.2(4.7) & = & 0\\-984.6 +983.2 & = & 0\\-985 +983 & = & 0\\0&=&0\end{array}$

The second term has only two significant figures because ΔT₂ has only two.

It agrees to two significant figures.

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