Question

0.5000 kg of water at 35.00 degrees Celsius is cooled, with the removal of 6.300 E4 J of heat. What is the final temperature of the water? Specific heat capacity of water is 4186 J/(kg Co).Remember to identity all of your data, write the equation, and show your work.

Answer 1

Answer:

65.1 °C

Explanation:

m = mass of the water = 0.5 kg

$T_{i}$ = initial temperature of water = 35.00 °C

$T_{f}$ = final temperature of water

c = specific heat of water = 4186 J/(kg °C)

Q = Amount of heat removed from water = 6.3 x 10⁴ J

Amount of heat removed from water is given as

Q = m c ($T_{f}$ - $T_{i}$)

Inserting the values

6.3 x 10⁴ = (0.5) (4186) ($T_{f}$ - 35.00)

$T_{f}$ = 65.1 °C

Answer 2

Answer : The final temperature of the water is $65.10^oC$

Explanation :

Formula used :

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat released = $6.300\times 10^{4}J$

m = mass of water = 0.5000 kg

c = specific heat of water = $4186J/kg^oC$

$T_{final}$ = final temperature = ?

$T_{initial}$ = initial temperature = $35.00^oC$

Now put all the given values in the above formula, we get:

$6.300\times 10^{4}J=(0.5000kg)\times (4186J/kg^oC)\times (T_{final}-35.00)^oC$

$T_{final}=65.10^oC$

Therefore, the final temperature of the water is $65.10^oC$