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

Approximately
`41.2\; {\rm ^(\circ) C}`.

Step-by-step 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)} * 50\; {\rm g} * (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)} * 25.0\; {\rm g} * (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}`.