Question

HELP FAST 100 PTS Calculate the amount of heat needed to raise the temperature of 96 g of water vapor from 124 °C to 158 °C. must provide explanation

Answer 1

`\Large \boxed{\sf 13600 \ J}`

Step-by-step explanation:

Use formula

`\displaystyle \sf Heat \ (J)=mass \ (g) * specific \ heat \ capacity \ (Jg^(-1)\°C^(-1)) * change \ in \ temperature \ (\°C)`

Specific heat capacity of water is 4.18 J/(g °C)

Substitute the values in formula and evaluate

`\displaystyle \sf Heat \ (J)=96 \ g * 4.18 \ Jg^(-1)\°C^(-1) * (158\°C-124 \°C)`

`\displaystyle Q=96 * 4.18 * (158-124 )=13643.52`

Answer 2

`\huge\boxed{\sf Q = 13.7\ Joules}`

Step-by-step explanation:

Given Data:

Mass = m = 96 g = 0.096 kg

`T_1` = 124 °C

`T_2` = 158 °C

Change in Temp. = ΔT = 158 - 124 = 34 °C

Specific Heat Constant = c = 4.186 J/g °C

Required:

Specific Heat Capacity = Q = ?

Formula:

Q = mcΔT

Solution:

Q = (0.096)(4.186)(34)

Q = 13.7 Joules

`\rule[225]{225}{2}`

Hope this helped!

~AH1807