How do you determine the mass of a water sample that is heated from an initial temperature of 25 deg C to a final temperature of 100 deg C following the addition of 1200 J of heat energy?

Question

Caleb Adams · Accepted Answer

Here's how I would do it.

The formula for the amount of heat absorbed by a substance is #color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "# where #q# is the quantity of heat #m# is the mass of the substance #c# is the specific heat capacity of the material #ΔT# is the temperature change You can rearrange the formula to calculate the mass: #m = q/(cΔT)# In your problem, #q = "1200 J"# #c = "4.184 J·°C"^"-1""g"^"-1"# (the specific heat capacity of water) #ΔT = T_f - T_i = "100 °C - 25 °C" = "75 °C"# ∴ #m = (1200 color(red)(cancel(color(black)("J"))))/(4.184 color(red)(cancel(color(black)("J·°C"^"-1")))"g"^"-1" × 75 color(red)(cancel(color(black)("°C")))) = "3.8 g"#

Natalie Anton · Answer

undefined You can determine the mass of the water sample using the formula:

$ q = mcΔT $

Where:
- $ q $ is the heat energy (in joules)
- $ m $ is the mass of the water (in grams)
- $ c $ is the specific heat capacity of water (4.18 J/g°C)
- $ ΔT $ is the change in temperature (in °C)

First, calculate the change in temperature:

$ ΔT = T_f - T_i $
$ ΔT = 100°C - 25°C = 75°C $

Then, rearrange the formula to solve for mass:

$ m = \frac{q}{cΔT} $

Substitute the given values:

$ m = \frac{1200 J}{4.18 J/g°C * 75°C} $
$ m ≈ 4.83 g $

So, the mass of the water sample is approximately 4.83 grams.