Question

A pan containing 30 grams of water was allowed to cool from a temperature of 90.0 °C. If the amount of heat released is 1,500 joules, what is the approximate final temperature of the water?
76 °C
78 °C
81 °C
82 °C

Answer 1

Answer:

Final temperature =78°C

Explanation:

The amount of heat lost is calculated using the formula for calculating the enthalpy change: mCΔT C, the specific heat capacity for for water is 4.186J/gK. The mass of water is 30 grams.

1500J= 30g×4.186J/gK×ΔT

ΔT=1500J/(30×4.186J/gK)

=11.94K

final temperature=(90-12)°C

=78°C

Answer 2

Answer:

Option B, 78.05 °C is the approximate final temperature of the water

Explanation:

Heat released$= m * c* (T_{2} - T_{1})$

Where "m" stands for mass of any substance

"c " stands for specific heat of water  is equal to 4.186 joule/gram °C

$T_{2}$ stands for final temperature

$T_{1}$ stands for initial temperature

Substituting the available values in above equation, we get -

$1500 = 30 * 4.186 * (90 - T_{1} )\\T_{1} = 90 - 11.94\\T_{1} = 78.05$

78.05 °C is the approximate final temperature of the water

Thus, option B is correct.