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Sagot :
Answer:
Approximately [tex]0.219\; {\rm kg}[/tex].
Explanation:
When the temperature of an object of specific heat capacity [tex]c[/tex] and mass [tex]m[/tex] changes by [tex]\Delta T[/tex], the associated energy change would be:
[tex]Q = c\, m\, \Delta T[/tex].
Rearranging this equation would give an expression for the mass of the object given the energy change, specific heat, and temperature change:
[tex]\displaystyle m = \frac{Q}{(c)\, (\Delta T)}[/tex].
In this question:
- Temperature change of the block of tin ([tex]m= 1.90\; {\rm kg}[/tex]): [tex]\Delta T = (150 - 64.0)\; {\rm K}[/tex].
- Temperature change of the water added to the container: [tex]\Delta T = (64.0 - 26.0)\; {\rm K}[/tex]
Apply the following steps to find the mass of water added:
- Find the energy that the block of tin released using the equation [tex]Q = c\, m\, \Delta T[/tex].
- Since energy conserved and the container is insulated, the energy water in the container absorbed would be equal to the energy the tin block released.
- Apply the following equation the find the mass of the water added:
[tex]\displaystyle m = \frac{Q}{(c)\, (\Delta T)}[/tex].
Look up the specific heat capacity of tin and water:
- Tin: approximately [tex]213\; {\rm J\cdot kg^{-1} \cdot K^{-1}}[/tex].
- Water: approximately [tex]4\, 186\; {\rm J\cdot kg \cdot K^{-1}}[/tex].
The energy that the block of tin released would be:
[tex]\begin{aligned} Q &= c\, m\, \Delta T \\ &= (213\; {\rm J\cdot kg^{-1} \cdot K^{-1}})\, (1.90\; {\rm kg})\, ((150 - 64.0)\; {\rm K}) \\ &\approx 34\, 800\; {\rm J}\end{aligned}[/tex].
Since the container is insulated, under the assumptions, energy water in the container absorbed would be equal to the energy the block of tin released: [tex]34\, 800\; {\rm J}[/tex]. Given that the temperature change of the water is [tex]\Delta T = (64.0 - 26.0)\; {\rm K}[/tex] and that the specific heat of water is approximately [tex]4\, 186\; {\rm J\cdot kg \cdot K^{-1}}[/tex], the mass of water added to the container would be:
[tex]\begin{aligned} m &= \frac{Q}{(c)\, (\Delta T)} \\ &\approx \frac{34\, 800\; {\rm J}}{(4\, 186\; {\rm J\cdot kg \cdot K^{-1}})\, ((64.0 - 26.0)\; {\rm K})} \\ &\approx 0.219\; {\rm kg}\end{aligned}[/tex].
In other words, the mass of water added to the container should be approximately [tex]0.219\; {\rm kg}[/tex].
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