Respuesta :
first find the enthalpy of fusion by 8.5gm* 333.5j/k=2834j
now find the specific heat of water by 4.186*2834/255=46.5k
this is the rquired temperature
now find the specific heat of water by 4.186*2834/255=46.5k
this is the rquired temperature
The temperature change in water upon the complete melting of 8.5 g of ice that is placed in 255 g of water is [tex]\boxed{2.66{\text{ }}^\circ {\text{C}}}[/tex].
Further explanation:
Latent heat of fusion:
When substance changes from solid state to liquid state, the amount of heat absorbed in this process without change in its temperature is calledlatent heat of fusion.
The expression for absorbed heat from the latent heat of fusion is as follows:
[tex]{\text{q}} = {\text{n}}{{\text{H}}_{\text{f}}}[/tex] …… (1)
Here,
q is heat absorbed.
n is number of moles of substance.
[tex]{{\text{H}}_{\text{f}}}[/tex] is latent heat of fusion.
The moles of water can be calculated by following formula:
[tex]{\text{Moles of water}} = \dfrac{{{\text{Mass of water}}}}{{{\text{Molar mass of water}}}}[/tex] …… (2)
Substitute 8.5 g for mass of water and 18.0152 g/mol for molar mass of water in equation (2).
[tex]\begin{aligned} {\text{Moles of water}} &= \left( {\frac{{{\text{8}}{\text{.5 g}}}}{{{\text{18}}{\text{.0152 g/mol}}}}} \right) \\ &= 0.4718{\text{ mol}} \\ \end{aligned}[/tex]
Substitute 0.4178 mol for n and 6.02 kJ/mol for [tex]{{\text{H}}_{\text{f}}}[/tex] in equation (1) to calculate heat absorbed by ice.
[tex]\begin{aligned} {\text{q}} &= \left( {{\text{0}}{\text{.4718 mol}}} \right)\left( {\frac{{{\text{6}}{\text{.02}}\;{\text{kJ}}}}{{{\text{1 mol}}}}} \right)\left( {\frac{{{{10}^3}{\text{ J}}}}{{1{\text{ kJ}}}}} \right) \\ & = 2.84 \times {10^3}{\text{ J}} \\ \end{aligned}[/tex]
Hence heat supplied by water for melting of ice becomes [tex]- 2.84 \times {10^3}{\text{ J}}[/tex].
The formula to calculate the heat energy of the solution is as follows:
[tex]{\text{q}} = \text{mc}\Delta T}}[/tex] …… (3)
Where,
q is amount of heat transferred.
m is mass of the solution.
c is specific heat.
[tex]{{\Delta T}}[/tex] is change in temperature.
Rearrange equation (3) to calculate .
[tex]{{\Delta T}} = \frac{{\text{q}}}{{{\text{mc}}}}[/tex] …… (4)
Substitute 255 g for m, [tex]- 2.84 \times {10^3}{\text{ J}}[/tex] for q and [tex]4.18\;{\text{J/g}}\;{\text{^\circ C}}[/tex] for c in equation (4).
[tex]\begin{aligned} {t{\Delta T}} &= \dfrac{{ - 2.84 \times {{10}^3}{\text{ J}}}}{{\left( {255{\text{ g}}} \right)\left( {4.186{\text{ J/g }}^\circ {\text{C}}} \right)}} \\ &= - 2.66{\text{ }}^\circ {\text{C}} \\ \end{aligned}[/tex]
Learn more:
- The difference between heat and temperature: https://brainly.com/question/914750
- Determine the process by which water enters into the atmosphere: https://brainly.com/question/2037060
Answer details:
Grade: Senior School
Subject: Chemistry
Chapter: Thermodynamics
Keywords: latent heat of fusion, q, Hf, 6.02 kJ/mol, 255 g, water, moles of water, molar mass of water, 18.0152 g/mol, 8.55 g.
