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Angelina_Jolie [31]
3 years ago
9

How much heat is required to heat 5.25 g of water h2o from 5.5?

Chemistry
1 answer:
Evgen [1.6K]3 years ago
5 0
Heat = mass * heat capacity of water * change in temperature mass = 5.25 g heat capacity of water = 4.186 joule/gram °C Change in temperature = 62.8°C - 5.3°C = 57.5 °C Plug in the values heat = 5.25 g * 4.186 joule/gram °C * 57.5 °C = 1263.6 J Rounded to two three significant figures, it is 1260 J of energy needed. In terms of calories, the heat capacity of water is 1 calorie/gram °C. So do the plugging in all over again. mass = 5.25 g heat capacity of water = 1 calorie/gram °C Change in temperature = 62.8°C - 5.3°C = 57.5 °C heat = 5.25 g * 1 calorie/gram °C * 57.5 °C = 301.9 calories Rounded to 3 significant figures, it is 302 calories Q=SM∆T=4.18*5.25*(62.8-4.3)=1280 J 1280 J * (1 cal/4.18 J) = 307 cal
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A chemist prepares a solution of sodium chloride(NaCl) by measuring out 25.4g of sodium chloride into a 100ml volumetric flask a
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6 0
3 years ago
calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
3 years ago
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