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In the lab, you submerge 100 g of 40°C nails in 200 g of 20°C water. (The specific heat of iron is 0.12 cal/g # °C.) Equate the heat gained by the water to the heat lost by the nails, and show that the final temperature of the water is about 21°C.

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Respuesta :

Answer:

Final temperature of the mixture becomes

[tex]T = 21.13^oC[/tex]

Explanation:

Here we know that heat given by the nail is equal to the heat absorbed by water

So here we know that heat to change the temperature is given as

[tex]Q = ms\Delta T[/tex]

so by equating the heat we have

[tex]m_{iron}s_{iron}\Delta T = m_{water}s_{water}\Delta T[/tex]

now we have

[tex]100 (0.12) (40 - T) = 200 (1) (T - 20)[/tex]

[tex]4.8 - 0.12 T = 2T - 40[/tex]

[tex]44.8 = 2.12 T[/tex]

[tex]T = 21.13^oC[/tex]

fichoh

Using the quantity of heat relation, the final temperature of water is [tex] 21.13°C [/tex]

Given the Parameters :

  • Mass of iron = 100g

  • Mass of water = 200g

  • Specific heat of iron, S = 0.12 cal/gram

  • Specific heat of water = 1 cal/gram

  • Change in Temperature of iron, △T = (40 - T)

  • Change in Temperature of water , △T = (T - 20)

Recall :

Heat lost by iron = Heat gained by water

[tex]M_{iron}S_{iron}\Delta T = M_{water}S_{water}\Delta T[/tex]

[tex] 100 \times 0.12 \times (40 - T) = 200 \times 1 \times (T-20)[/tex]

[tex] 12 \times (40 - T) = 200 \times (T-20)[/tex]

[tex] 480 - 12T = 200T - 4000 [/tex]

Collect like terms

[tex]- 12T - 200 T = - 4000 - 480 [/tex]

[tex] - 212T = - 4480 [/tex]

[tex] T = \frac{4480}{212} [/tex]

[tex] T = 21.13°C [/tex]

Therefore the final temperature, [tex] T \: is \: 21.13°C [/tex]

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