Question

A 75.0 g sample of granite initially at 92.5oC is immersed into 60.0 g of water initially at 22.2oC. Determine the final temperature (in oC) when they reach thermal equilibrium. Assume no heat loss to the surroundings.

Specific heats: granite = 0.790 J/(goC), water = 4.184 J/goC

Answer 1

`\large \boxed{35.6 \,^(\circ)\text{C}}`

Step-by-step explanation:

There are two heat transfers involved: the heat lost by the granite 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 granite be Component 1 and the water be Component 2.

Data:

For the granite:

`m_(1) =\text{75.0 g; }T_(i) = 92.5 ^(\circ)\text{C; }\\C_(1) = 0.790 \text{ J$^(\circ)$C$^(-1)$g$^(-1)$}`

For the water:

`m_(2) =\text{60.0 g; }T_(i) = 22.2 ^(\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 granite + heat gained by water} & = & 0\\m_(1)C_(1)\Delta T_(1) + m_(2)C_(2)\Delta T_(2) & = & 0\\\text{75.0 g}* 0.790 \text{ J$^(\circ)$C$^(-1)$g$^(-1)$} *\Delta T_(1) + \text{60.0 g} * 4.184 \text{ J$^(\circ)$C$^(-1)$g$^(-1)$}\Delta * T_(2) & = & 0\\59.25\Delta T_(1) + 251.0\Delta T_(2) & = & 0\\\end{array}`

(b) Final temperature of water

`\Delta T_(1) = T_{\text{f}} - 92.5 ^(\circ)\text{C}\\\Delta T_(2) = T_{\text{f}} - 22.2 ^(\circ)\text{C}`

`\begin{array}{rcl}59.25(T_{\text{f}} - 92.5 \, ^(\circ)\text{C}) + 251.0(T_{\text{f}} - 22.2 \, ^(\circ)\text{C}) & = & 0\\59.25T_{\text{f}} - 5481 \, ^(\circ)\text{C} + 251.0T_{\text{f}} - 5572 \, ^(\circ)\text{C} & = & 0\\310.2T_{\text{f}} - 11052\, ^(\circ)\text{C} & = & 0\\310.2T_{\text{f}} & = & 11052 \, ^(\circ)\text{C}\\T_{\text{f}}& = & \mathbf{35.6 \, ^(\circ)}\textbf{C}\\\end{array}\\\text{The final temperature is $\large \boxed{\mathbf{35.6 \,^(\circ)}\textbf{C}}$}`