2834.75 = 255 (4.16) (T)
T = 2.67K
The law of conservation of energy is an important concept for solving this problem. The energy released is enough for the energy absorbed.
Ice undergoes a heat of transformation. This means there is a phase change but not temperature and therefore energy is resolved by (fusion enthalpy) (mass) = 333.5J / g) (8.5g). = 2834.75J.
This is enough for the energy released by water. Energy calculated by (mass) (specific water heat) (temperature change) = (255g) (4.16J / gK) (T)
In physics, the law of conservation of energy states that the amount of energy from a closed system does not change - it will remain the same. That energy cannot be created or destroyed by humans, but it can change from one form of energy to another. For example, chemical energy can be converted into kinetic energy in dynamite explosions.
The most common definition of energy is a work that can be carried out by certain forces (gravity, electromagnetic, etc.).
Energy is a conserved quantity. Because of various forces, such as gravity, electricity, heat and so on, energy has various forms.
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Details
Grade: High School
Subject: Chemistry
Keyword: conservation, law, energy
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:
When the physical state of substance changes from solid to liquid, the amount of heat absorbed in this process for complete conversion of solid into liquid without altering the temperature is known as the latent heat of fusion.
The formula to calculate heat absorbed from the latent heat of fusion is as follows:
[tex]{\text{q}} = {\text{n}}{{\text{H}}_{\text{f}}}[/tex] …… (1)
Where,
q is heat absorbed.
n is the number of moles of substance.
[tex]{{\text{H}}_{\text{f}}}[/tex] is latent heat of fusion.
The formula to calculate the moles of water is as follows:
[tex]{\text{Moles of water}} = \dfrac{{{\text{Given mass of water}}}}{{{\text{Molar mass of water}}}}[/tex] …… (2)
The mass of water is 8.5 g.
The molar mass of water is 18.0152 g/mol.
Substitute these values in equation (2).
[tex]\begin{aligned} {\text{Moles of water}} &= \left( {{\text{8}}{\text{.5 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{18}}{\text{.0152 g}}}}} \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).
[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 the amount of heat transferred.
m is the mass of the solution.
c is the specific heat.
[tex]{\Delta T}}[/tex] is the change in temperature.
Rearrange equation (3) to calculate [tex]{{\Delta T}}[/tex] .
[tex]{{\Delta T}} = \dfrac{{\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} {{\Delta T}} &= \frac{{ - 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]
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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.
