1 answer

A silver block, initially at 58.2 ?C, is submerged into 100.0 g of water at 25.0 ?C, in an insulated

Question:

A silver block, initially at 58.2 ?C, is submerged into 100.0 g of water at 25.0 ?C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 27.6 ?C.

-What is the mass of the silver block?


Answers

Concepts and reason

The given problem is based on the principle of heat exchange between two systems. The mass of the silver block needs to be calculated which is submerged into the given mass of water. The mixture reached thermal equilibrium at temperature 27.6C27.6^\circ {\rm{C}} therefore, by equating the heat loss by one system and gained by the other system, calculate the mass of silver block.

Fundamentals

Heat capacity can be defined as the amount of heat released or absorbed to change the temperature of the compound. Mathematically, it can be represented as follow:

q=m.c.ΔTq = m{\rm{.c}}{\rm{.}}\Delta T

Here,

q=Amountofheatm=Massofsubstancec=specificheatcapacityΔT=temperaturechange\begin{array}{c}\\q = {\rm{Amount}}\,{\rm{of}}\,{\rm{heat}}\\\\m = {\rm{Mass of substance}}\\\\{\rm{c}} = {\rm{specific}}\,{\rm{heat}}\,{\rm{capacity}}\\\\\Delta T = {\rm{temperature}}\,{\rm{change}}\\\end{array}

Use the following equation to calculate the heat gain by 100g100\,{\rm{g}} of water when the temperature of water is raised from 25C25\,^\circ {\rm{C}} to 27.6C27.6\,^\circ {\rm{C}} as shown below:

q1(Heatgainbywatersystem)=m.c.ΔT=100.0g×4.186J/gC×(27.6C25C)=1088.36J\begin{array}{c}\\{q_1}\,\left( {{\rm{Heat}}\,{\rm{gain}}\,{\rm{by}}\,{\rm{water}}\,{\rm{system}}} \right) = m{\rm{.c}}{\rm{.}}\Delta T\\\\ = 100.0\,{\rm{g}} \times 4.186{\rm{ J}}/{\rm{g}}\,^\circ {\rm{C}} \times \left( {27.6\,^\circ {\rm{C}} - 25\,^\circ {\rm{C}}} \right)\\\\ = 1088.36{\rm{ J}}\\\end{array}

Calculate the mass of silver block as shown below:

q2(Heatlossbysilversystem)=msilver.csilver.ΔT1088.36J=msilver×0.239J/gC×(27.2C58.2C)msilver=1088.36J7.41J/g=148.8g\begin{array}{c}\\ - {q_2}\left( {{\rm{Heat}}\,{\rm{loss}}\,{\rm{by}}\,{\rm{silver}}\,{\rm{system}}} \right) = {m_{silver}}{\rm{.}}{{\rm{c}}_{silver}}{\rm{.}}\Delta T\\\\ - 1088.36{\rm{ J}} = {m_{silver}} \times 0.239{\rm{ J}}/{\rm{g}}\,^\circ {\rm{C}} \times \left( {27.2\,^\circ {\rm{C}} - 58.2\,^\circ {\rm{C}}} \right)\\\\{m_{silver}} = \frac{{1088.36{\rm{ J}}}}{{7.41\,{\rm{J}}/{\rm{g}}}} = 148.8\,{\rm{g}}\\\end{array}

Ans:

Hence, the mass of the silver block is calculated as 148.8g148.8\,{\rm{g}} .

.

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