Physics, 01.04.2020 06:27 ingle75

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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Answer from: xxgissellexx

Using the quantity of heat relation, the final temperature of water is 21.13°C

Given the Parameters :

Mass of iron = 100g Mass of water = 200g Specific heat of iron, S = 0.12 cal/gramSpecific heat of water = 1 cal/gramChange in Temperature of iron, △T = (40 - T) Change in Temperature of water , △T = (T - 20)

Recall :

Heat lost by iron = Heat gained by water

$M_{iron}S_{iron}\Delta T = M_{water}S_{water}\Delta T$

$100 \times 0.12 \times (40 - T) = 200 \times 1 \times (T-20)$

$12 \times (40 - T) = 200 \times (T-20)$

480 - 12T = 200T - 4000

Collect like terms

- 12T - 200 T = - 4000 - 480

- 212T = - 4480

$T = \frac{4480}{212}$

T = 21.13°C

Therefore the final temperature, $T \: is \: 21.13°C$

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Answer from: whitewesley124

Final temperature of the mixture becomes

T = 21.13^oC

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

$Q = ms\Delta T$

so by equating the heat we have

$m_{iron}s_{iron}\Delta T = m_{water}s_{water}\Delta T$

now we have

100 (0.12) (40 - T) = 200 (1) (T - 20)

4.8 - 0.12 T = 2T - 40

44.8 = 2.12 T

T = 21.13^oC

Answer from: Quest

The answer is a.first and third quarters

Answer from: Quest

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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...

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