Question

What would be the final temperature of the system if 21.2 g of lead at 113 ◦C is dropped into 14.8 g of water at 28.9 ◦C in an insulated container? The specific heat of lead is 0.128 J/g◦C.

Answer 1

The final temperature of the system is 32.4 °C.

heat lost by Pb + heat gained by water = 0

`q_(1) +q_(2) = 0`

`m_(1)C_(1)\Delta T_(1) + m_(2)C_(2)\Delta T_(2) = 0`

`q_(1)= m_(1)C_(1)\Delta T_(1) = 21.2 \:\text{g}* 0.128\: \text{J}^(\circ)\text{C}^(-1)\text{g}* (T_(f) -113\: ^(\circ)\text{C}) = 2.713T_(f)\: \text{J}^(\circ)\text{C}^(-1)\: -\: 306.6\:\text{J}\cdot\text{g}^(-1)\\`

`q_(2)= m_(2)C_(2)\Delta T_(2) = 14.8 \:\text{g}* 4.184\: \text{J}\cdot^(\circ)\text{C}^(-1)\text{g}^(-1)* (T_(f) -28.9\: ^(\circ)\text{C}) = 61.9\: \Delta T_(f) \: \text{J}^(\circ)\text{C} \:- 1790 \: \text{J}\\`

Adding
`q_(1)` and
`q_(2)` and combinibg like terms, we get

`168.0\text{T}_(f)\: ^(\circ)\text{C} ^(-1)\:- 2096 = 0]\\`

`T_(f) = \frac{2096}{68.0 \: ^(\circ)\text{C}^(-1)} = 32.4 ^(\circ)\text{C}\\`