Question

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.
PLEASE NEED BY TODAY

Answer 1

Answer:

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

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

$4.8 - 0.12 T = 2T - 40$

$44.8 = 2.12 T$

$T = 21.13^oC$