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

Answer:

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

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

Answer 2

Answer:

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

Explanation:

Use formula

$\displaystyle \sf Heat \ (J)=mass \ (g) \times specific \ heat \ capacity \ (Jg^{-1}\°C^{-1}) \times change \ in \ temperature \ (\°C)$

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

Substitute the values in formula and evaluate

$\displaystyle \sf Heat \ (J)=96 \ g \times 4.18 \ Jg^{-1}\°C^{-1} \times (158\°C-124 \°C)$

$\displaystyle Q=96 \times 4.18 \times (158-124 )=13643.52$