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 s

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

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niczorrrr 6 months 2021-07-20T03:01:58+00:00 1 Answers 18 views 0

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    2021-07-20T03:03:14+00:00

    Answer:

    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\!.)

    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} \times 50\; \rm g \times (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} \times 250\; \rm g \times (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}} \times (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}} \times (x\; \rm g) \times (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.

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