The burning of a sweet potato sample generates 102.5 kJ of heat. This heat was used to raise the temperature of 300 grams of water from its initial temperature of 15.5 degrees Celsius. Considering the specific heat of water is 4.184 J/g-degrees-C, what is the final temperature of the water

97.16°C

Explanation:

Step 1:

Data obtained from the question. This includes:

Heat (Q) = 102.5 kJ = 102.5 x 1000 = 102500J

Mass (M) = 300 g

Initial temperature (T1) = 15.5°C

Specific heat capacity (C) = 4.184 J/g°C

Final temperature (T2) =?

Change in temperatures (ΔT) = T2 - T1 = T2 - 15.5

Step 2:

Determination of the final temperature.

Applying the following equation:

Q = MCΔT

We can easily calculate the value of the final temperature as follow:

Q = MCΔT

Q = MC(T2 - T1)

102500 = 300 x 4.184 (T2 - 15.5)

102500 = 1255.2 x (T2 - 15.5)

Divide both side by 1255.2

(T2 - 15.5) = 102500/1255.2

T2 - 15.5 = 81.66

Collect like terms

T2 = 81.66 + 15.5

T2 = 97.16°C

Therefore, the final temperature of the water is 97.16°C

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.