Question

HELP FAST 100 PTSCalculate the amount of heat needed to convert 100.0 g of liquid water at 25 °C to water at 100 °C.

Answer 1

Answer:

$\Large \boxed{\sf 31400\ J}$

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

$\displaystyle \sf Heat \ (J)=100.0 \ g \times 4.18 \ Jg^{-1}\°C^{-1} \times (100\°C-25\°C)$

$\displaystyle Q=100.0 \times 4.18 \times (100-25 )=31350$

Answer 2

Answer:

31,380 Joules

Explanation:

Given Data:

Mass = m = 100 g

Temperature 1 = = 25 °C

Temperature 2 = = 100 °C

Specific Heat Constant = c = 4.184

Change in Temp. = ΔT = 100 - 25 = 75 °C

Required:

Heat = Q = ?

Formula:

Q = mcΔT

Solution:

Q = (100)(4.184)(75)

Q = 31, 380 Joules

Hope this helped!

~AH1807