#1.2 xx 10^5# #J# of thermal energy are added to a sample of water and its temperature changes from #229# #K# to #243# #K#. What is the mass of the water?

Question

Peyton Adams · Accepted Answer

#Delta H = mCDeltaT#

Rearranging:

#m = (Delta H)/(CDeltaT)=(1.2xx10^5)/(4181xx(243-229))=2.1# #kg# We use the equation #Delta H = mCDeltaT# where #m# is the mass #(kg)#, #T# is the temperature #(K)#, #H# is the thermal energy #(J)# and #C# is the specific heat #(Jkg^-1K^-1)#. #Delta# just means 'the change in'. We know that for water, #C=4181# #Jkg^-1K^-1#. (Sometimes quoted as #4.2# #Jg^-1K^-1#, but it's better to work in SI units.)

Jeremiah Adams · Answer

undefined To find the mass of the water, we can use the formula:

$$ Q = mc\Delta T $$

where:
- $ Q $ is the thermal energy added (1.2 x 10^5 J),
- $ m $ is the mass of the water (in kg),
- $ c $ is the specific heat capacity of water (approximately 4.18 J/g·K), and
- $ \Delta T $ is the change in temperature (final temperature - initial temperature).

Rearranging the formula to solve for $ m $:

$$ m = \frac{Q}{c\Delta T} $$

Substituting the given values:

$$ m = \frac{1.2 \times 10^5 \, \text{J}}{4.18 \, \text{J/g·K} \times (243 \, \text{K} - 229 \, \text{K})} $$

$$ m \approx \frac{1.2 \times 10^5 \, \text{J}}{4.18 \, \text{J/g·K} \times 14 \, \text{K}} $$

$$ m \approx \frac{1.2 \times 10^5 \, \text{J}}{58.52 \, \text{J/g}} $$

$$ m \approx 2050.07 \, \text{g} $$

$$ m \approx 2.05 \, \text{kg} $$