An insulated container is used to hold
4
6
.
2
g of water at
3
2
.
5

\
deg C
.
A sample of copper weighing
1
4
.
2
g is placed in a dry test tube and heated for
3
0
minutes in a boiling water bath at
1
0
0
.
0
\
deg C
.
The heated test tube is carefully removed from the water bath with laboratory tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is
0
.
3
8
5
J
/
(
g
·
\
deg C
)
,
calculate the maximum temperature of the water in the insulated container after the copper metal is added.

Question

An insulated container is used to hold 
4
6
.
2
 ﻿ g of water at 
3
2
.
5
 
\
deg C
.
 ﻿A sample of copper weighing 
1
4
.
2
 ﻿ g is placed in a dry test tube and heated for 
3
0
 ﻿minutes in a boiling water bath at 
1
0
0
.
0
\
deg C
.
 ﻿The heated test tube is carefully removed from the water bath with laboratory tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is 
0
.
3
8
5
 ﻿J
/
(
g
·
\
deg C
)
,
 ﻿calculate the maximum temperature of the water in the insulated container after the copper metal is added.

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Calculate the heat absorbed by water**: To find the maximum temperature of the water, we need to calculate the heat transferred from the copper to the water. The heat absorbed by the water ($q_{water}$) can be calculated using the formula $q = mc\Delta T$, where $m$ is the mass of the water, $c$ is the specific heat capacity of water (which is $4.18 \, ext{J/g}^\circ ext{C}$), and $\Delta T$ is the change in temperature. ‖ ‖ 2 ⋮ **Calculate the heat lost by copper**: Similarly, the heat lost by the copper ($q_{copper}$) as it cools down can be calculated using the same formula $q = mc\Delta T$, where $m$ is the mass of the copper, $c$ is the specific heat capacity of copper ($0.385 \, ext{J/g}^\circ ext{C}$), and $\Delta T$ is the change in temperature of the copper. Since the copper is initially at $100.0^\circ ext{C}$ and the final temperature of the copper will be the same as the final temperature of the water, we can express $\Delta T$ for copper as $(100.0 - T_{final})$. ‖ ‖ 3 ⋮ **Apply the law of conservation of energy**: According to the law of conservation of energy, the heat lost by the copper will be equal to the heat gained by the water. Therefore, we can set $q_{copper}$ equal to $q_{water}$ and solve for the final temperature ($T_{final}$). ‖ ‖ 4 ⋮ **Solve for the final temperature**: By setting the heat lost by copper equal to the heat gained by water, we can solve for $T_{final}$. The equation is $m_{copper}c_{copper}(100.0 - T_{final}) = m_{water}c_{water}(T_{final} - 32.5)$. Plugging in the values, we get $14.2 imes 0.385 imes (100.0 - T_{final}) = 462 imes 4.18 imes (T_{final} - 32.5)$. Solving for $T_{final}$ will give us the maximum temperature of the water. ‖ ∻Answer∻ ⚹ [Insert final answer here] ⚹ ∻Key Concept∻ ⚹ Heat transfer and the law of conservation of energy ⚹ ∻Explanation∻ ⚹ The maximum temperature of the water is determined by the heat transferred from the copper to the water, which is calculated using the specific heat capacities and masses of the substances involved. The law of conservation of energy states that the heat lost by the copper will be equal to the heat gained by the water. ⚹◊What is the equation that can be used to calculate the final temperature of the water in the insulated container after adding the heated copper metal?⍭ Generate me a similar question◊

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