Question

burning propane heats a 712.3 g water sample from 4 to 29°c. how much energy was absorbed by the water in joules? input the numeric value here. definitions and standards: 1000 j = 1 kj, d = delta, dt = change in temperature (°c), dhvap = heat of vaporization (j/g), dhfus = heat of fusion (j/g), s = specific heat (j/g°c), s water = 4.184 j/g°c, s steam = 2.02 j/g°c, s ice = 2.05 j/g°c, dhfush2o = 334 j/g, dhvaph2o = 2260 j/g, ccalorimetry = heat capacity (kj/°c) formula bank: 1. q = m·c·δt q = m·s·δt 2. q = c·δt 3. qvap = m·δhvap 4. qfus = m·δhfus 5. δh°rxn = σ(δhfprod) - σ(dhfreact) 6. δu = q + w δe = q + w 7. w = δu - q w = δe - q 8. w = -p·δv these problems are solved by dimensional analysis, formulas, and simple math. be sure to show the formula on exams. you must show work on the exams, so practice here. practice showing the calculator answer and the sig figs answer. be sure to report the units for all values when showing your work. show work on the exams will be handed in or by uploading a jpg or pdf file (upload not required here). keys will be posted for exemplary practice problems.

Explanation:

Step1: Calculate temperature change

$\Delta T=T_{final}-T_{initial}=29 - 4=25^{\circ}C$

Step2: Use heat - absorption formula

The formula for heat absorbed by a substance is $q = m\times s\times\Delta T$. Given $m = 712.3g$, $s = 4.184J/g^{\circ}C$ (specific heat of water), and $\Delta T = 25^{\circ}C$.
Substitute the values into the formula: $q=712.3\times4.184\times25$
First, calculate $712.3\times4.184 = 712.3\times(4 + 0.184)=712.3\times4+712.3\times0.184=2849.2+131.0632 = 2980.2632$
Then, $q = 2980.2632\times25=74506.58J$

Answer:

$74507$ (rounded to the nearest whole number)