Question

A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperature of the system 50 C. ( specific heat water= 4200 J/Kg C , specific heat copper= 390 J/Kg C

Answer 1

Approximately
`13\; \rm g` of steam at
`100\; \rm ^\circ C` (assuming that the boiling point of water in this experiment is
`100\; \rm ^\circ C\!`.)

Step-by-step explanation:

Latent heat of condensation/evaporation of water:
`2260\; \rm J \cdot g^(-1)`.

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to
`\rm J \cdot g^(-1)`.

Specific heat of water:
`4.2\; \rm J \cdot g^(-1)\cdot \rm K^(-1)`.

Specific heat of copper:
`0.39\; \rm J \cdot g^(-1)\cdot K^(-1)`.

The temperature of this calorimeter and the
`250\; \rm g` of water that it initially contains increased from
`20\; \rm ^\circ C` to
`50\; \rm ^\circ C`. Calculate the amount of energy that would be absorbed:

`\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^(-1)\cdot K^(-1) * 50\; \rm g * (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J \end{aligned}`.

`\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^(-1)\cdot K^(-1) * 250\; \rm g * (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J \end{aligned}`.

Hence, it would take an extra
`585\; \rm J + 31500\; \rm J = 32085\; \rm J` of energy to increase the temperature of the calorimeter and the
`250\; \rm g` of water that it initially contains from
`20\; \rm ^\circ C` to
`50\; \rm ^\circ C`.

Assume that it would take
`x` grams of steam at
`100\; \rm ^\circ C` ensure that the equilibrium temperature of the system is
`50\; \rm ^\circ C`.

In other words,
`x\; \rm g` of steam at
`100\; \rm ^\circ C` would need to release
`32085\; \rm J` as it condenses (releases latent heat) and cools down to
`50\; \rm ^\circ C`.

Latent heat of condensation from
`x\; \rm g` of steam:
`2260\; {\rm J \cdot g^(-1)} * (x\; {\rm g}) = (2260\, x)\; \rm J`.

Energy released when that
`x\; {\rm g}` of water from the steam cools down from
`100\; \rm ^\circ C` to
`50\; \rm ^\circ C`:

`\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^(-1)\cdot K^(-1)} * (x\; \rm g) * (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J \end{aligned}`.

These two parts of energy should add up to
`32085\; \rm J`. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from
`20\; \rm ^\circ C` to
`50\; \rm ^\circ C`.

`(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J`.

Solve for
`x`:

`x \approx 13`.

Hence, it would take approximately
`13\; \rm g` of steam at
`100\; \rm ^\circ C` for the equilibrium temperature of the system to be
`50\; \rm ^\circ C`.