Question

if a sample of water releases 3251 j of energy as the temperature decreases from 25 to 17°c, what was the mass of the sample? input the numeric value here. definitions and standards: 1000 j = 1 kj, d = delta, dt = change in temperature (°c), dh_vap = heat of vaporization (j/g), dh_fus = 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, dh_fush2o = 334 j/g, dh_vaph2o = 2260 j/g, c_calorimetry = heat capacity (kj/°c) formula bank: 1. q = m·c·δt q = m·s·δt 2. q = c·δt 3. q_vap = m·δh_vap 4. q_fus = m·δh_fus 5. δh°_rxn = σ(δhf_prod) - σ(dhf_react) 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 the temperature change

$\Delta T=T_{final}-T_{initial}=17 - 25=- 8^{\circ}C$

Step2: Rearrange the formula $q = m\cdot s\cdot\Delta T$ to solve for $m$

$m=\frac{q}{s\cdot\Delta T}$
We know that $q = 3251\ J$, $s = 4.184\ J/g^{\circ}C$, and $\Delta T=-8^{\circ}C$. Substitute these values into the formula:
$m=\frac{3251\ J}{4.184\ J/g^{\circ}C\times(- 8^{\circ}C)}$
Since we are interested in the magnitude of mass (mass cannot be negative), we can use $|\Delta T| = 8^{\circ}C$
$m=\frac{3251}{4.184\times8}$
$4.184\times8 = 33.472$
$m=\frac{3251}{33.472}\approx97.1\ g$

Answer:

$97.1$