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

`\Large \boxed{\sf 31400\ 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)=100.0 \ g * 4.18 \ Jg^(-1)\°C^(-1) * (100\°C-25\°C)`

`\displaystyle Q=100.0 * 4.18 * (100-25 )=31350`

Answer 2

31,380 Joules

Step-by-step 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