Question

25.0g of iron is heated to 100.0 and then placed in 50.0 g of water in a insulated calorimeter. the initial temperature of the water is 38.00. the specific heat of water is 4.181j/g and the specific heat if iron is 0.45j/g. what is the final temp of the water and the iron?

Answer 1

Answer:

Approximately $41.2\; {\rm ^{\circ} C}$.

Explanation:

Let $t\; {\rm ^{\circ} C}$ be the final temperature of the water and the iron.

Temperature of the water would be increase by $(t - 38.00)\; {\rm ^{\circ} C}$.

Temperature of the iron would be reduced by $(100.0 - t)\; {\rm ^{\circ} C}$.

Let $c$ denote the specific heat of each material. Let $m$ denote the mass of the material. For a temperature change of $\Delta t$, the energy change involved would be:

$Q = c\, m \, \Delta t$.

The energy that the water need to absorb would be:

$\begin{aligned}& Q(\text{water, absorbed}) \\ =\; & c(\text{water}) \, m(\text{water})\, \Delta t (\text{water}) \\ =\; & 4.181\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 50\; {\rm g} \times (t - 38.00)\; {\rm ^{\circ} C} \\ =\; & (209.05\, t - 7943.9)\; {\rm J} \end{aligned}$.

The energy that the iron would need to release would be:

$\begin{aligned}& Q(\text{iron, released}) \\ =\; & c(\text{iron}) \, m(\text{iron})\, \Delta t (\text{iron}) \\ =\; & 0.45\; {\rm J \cdot g^{-1} \cdot K^{-1}} \times 25.0\; {\rm g} \times (100.0 - t)\; {\rm ^{\circ} C} \\ =\; & (1125 - 11.25 \, t)\; {\rm J} \end{aligned}$.

Since this calorimeter is insulated, the energy that the iron had released would be equal to the energy that the water had absorbed:

$Q(\text{water, absorbed}) = Q(\text{iron, released})$.

$209.05\, t - 7943.9 = 1125 - 11.25\, t$.

$t \approx 41.2$.

Thus, the final temperature of the water and the iron would be approximately $41.2\; {\rm ^{\circ} C}$.