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A 500-g chunk of an unknown metal, which has been in boiling water for several minutes, is...

A 500-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg ofwater at room temperature (20.0 C). After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reach a constant value of 22.0C.A) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specic heat of the metal? b) Whichis more useful for storing heat, this metal or an equal weight of water? Explain. c) What if the heat absorbed by the Styrofoam actually is not negligible. How wouldthe specic heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain
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Answer #1
Assuming that metal has attained a tempreture of100oC in boiling water

Heat lost by metal = heat gained by water in beaker

0.5c(100 - 22) = 1*4200*(22-20) where c isthe specific haet capacity of the metal

39c = 8400 or c= 215.4 J /kg oC

b)The metal as it carries large amount of heat in a small massand it does not loose heat in evaporation

c)if heat loss in beaker is not negligible, the error inc of metal would give the value of c too small.

hope this helps.god bless you.
answered by: Alice S.O.S
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Answer #2
working on the question.wait for a few minutes
answered by: Wesley
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Answer #3

General guidance

Concepts and reason

The concept required to solve the given problem is mirror formula and magnification.

For the first part, use the law of conservation of energy and the formula for the amount of heat lost by a substance to determine the specific heat of the unknown metal.

For the second part use the Newton\u2019s law of cooling to determine the best material for storing heat.

Fir the last part, use the law of conservation of energy to determine the error in the prediction of specific heat of the metal when the heat absorbed by Styrofoam is taken into consideration.

Fundamentals

Law of conservation of energy: According to this theorem the total energy of an isolated system remains constant. It is said to be conserved over a period of time.

Newton\u2019s law of cooling: It states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature (i.e. the temperature of its surroundings).

The heat lost to the surrounding can be calculated as,

Q=mC\u0394TQ = mC\\Delta T

Here, mm is the mass of the substance, CC is the specific heat, and \u0394T\\Delta T is the change in temperature.

First Step | All Steps | Answer Only

Step-by-step

Step 1 of 3

(a)

The heat lost to the surrounding can be calculated as,

Q=mC\u0394TQ = mC\\Delta T

Here, mm is the mass of the substance, CC is the specific heat and \u0394T\\Delta T is the change in temperature.

Now, the heat lost by the metal will be equal to the heat gained by water.

\u0394Qlost=\u0394Qgain\\Delta {Q_{{\\rm{lost}}}} = \\Delta {Q_{{\\rm{gain}}}}

Substitute mmetalCmetal\u0394Tmetal{m_{{\\rm{metal}}}}{C_{{\\rm{metal}}}}\\Delta {T_{{\\rm{metal}}}} for \u0394Qlost\\Delta {Q_{lost}} and mwaterCwater\u0394Twater{m_{{\\rm{water}}}}{C_{{\\rm{water}}}}\\Delta {T_{{\\rm{water}}}}for \u0394Qgain\\Delta {Q_{{\\rm{gain}}}}in the above equation.

mmetalCmetal\u0394Tmetal=mwaterCwater\u0394Twater{m_{{\\rm{metal}}}}{C_{{\\rm{metal}}}}\\Delta {T_{{\\rm{metal}}}} = {m_{{\\rm{water}}}}{C_{{\\rm{water}}}}\\Delta {T_{{\\rm{water}}}}

Rearrange the above equation.

Cmetal=mwaterCwater\u0394Twatermmetal\u0394Tmetal{C_{{\\rm{metal}}}} = \\frac{{{m_{{\\rm{water}}}}{C_{{\\rm{water}}}}\\Delta {T_{{\\rm{water}}}}}}{{{m_{{\\rm{metal}}}}\\Delta {T_{{\\rm{metal}}}}}} \u2026\u2026 (1)

The change in temperature of the unknown metal will be,

\u0394Tmetal=Tf(metal)\u2212Ti(metal)=(100\u2218C)\u2212(22\u2218C)=78\u2218C\\begin{array}{c}\\\\\\Delta {T_{{\\rm{metal}}}} = {T_{f\\left( {{\\rm{metal}}} \\right)}} - {T_{i\\left( {{\\rm{metal}}} \\right)}}\\\\\\\\ = \\left( {100{\\rm{ }}^\\circ {\\rm{C}}} \\right) - \\left( {22{\\rm{ }}^\\circ {\\rm{C}}} \\right)\\\\\\\\ = 78{\\rm{ }}^\\circ {\\rm{C}}\\\\\\end{array}

The change in temperature of the water will be,

\u0394Twater=Tf(water)\u2212Ti(water)=(22\u2218C)\u2212(20\u2218C)=2\u2218C\\begin{array}{c}\\\\\\Delta {T_{{\\rm{water}}}} = {T_{f\\left( {{\\rm{water}}} \\right)}} - {T_{i\\left( {{\\rm{water}}} \\right)}}\\\\\\\\ = \\left( {{\\rm{22 }}^\\circ {\\rm{C}}} \\right) - \\left( {20{\\rm{ }}^\\circ {\\rm{C}}} \\right)\\\\\\\\ = 2{\\rm{ }}^\\circ {\\rm{C}}\\\\\\end{array}

Substitute 2\u2218C2{\\rm{ }}^\\circ {\\rm{C}} for \u0394Twater\\Delta {T_{{\\rm{water}}}}, 78\u2218C78{\\rm{ }}^\\circ {\\rm{C}} for \u0394Tmetal\\Delta {T_{{\\rm{metal}}}}, 500g500{\\rm{ g}}for mmetal{m_{{\\rm{metal}}}}

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