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Question: A Block A is a 50g aluminum block originally at \[{90^ \circ }C\]. Block B is a 100-g aluminum block...

A Block A is a 50g aluminum block originally at 90∘C{90^ \circ }C. Block B is a 100-g aluminum block original at 45∘C{45^ \circ }C. The blocks are placed in two separate 1.0 liter containers of water that were originally at 20∘C{20^ \circ }C. When the systems reach thermal equilibrium, which aluminum block will have the higher final temperature?
(A) Block A
(B) Block B
(C) The Blocks will have same final temperature
(D) The answer depends on the specific heat of water

Explanation

Solution

Understand the given data. We can use the Energy formula, keeping specific heat as constant, since both are aluminum. Equate the energy formula for A and B to find the relation between the final temperatures.

Complete step by step solution:
Let’s understand the given data. It is given that Block A made of aluminum weighing 80 grams is originally at 90∘C{90^ \circ }C. Another Block made of aluminum, weighing 100g is originally at 45∘C{45^ \circ }C. Now the blocks have a similar specific heat, since it’s made of similar material.
When the blocks are placed inside the water container, the heat energy dispersed or the heat energy lost is assumed to be equal. Therefore QA=QB{Q_A} = {Q_B}.
We know that, the formula for Heat Energy Q, which consists of mass and temperature as,
Q=mcΔTQ = mc\Delta T, Where, m is mass of the block and c is specific heat of the material and T as change in temperature from initial to final.
Using this, we get
QA=mAcΔTA{Q_A} = {m_A}c\Delta {T_A}(For Block A)
⇒QA=50×c×(90−T)\Rightarrow {Q_A} = 50 \times c \times (90 - T)------ (1)
Now for Block B,
QB=mBcΔTB{Q_B} = {m_B}c\Delta {T_B}
⇒QB=100×c×(45−T)\Rightarrow {Q_B} = 100 \times c \times (45 - T)------- (2)
Now, we need to identify the heat energy dispersed by the water medium. This implies
QW=mWcWΔTW{Q_W} = {m_W}{c_W}\Delta {T_W}
⇒QW=mW×cW×(T−20)\Rightarrow {Q_W} = {m_W} \times {c_W} \times (T - 20)------- (3)
Now equate (1) and (3) for Block A immersed in a water container.
mAcAΔTW=mWcWΔTW{m_A}{c_A}\Delta {T_W} = {m_W}{c_W}\Delta {T_W}
⇒50×c×(90−TA′)=mW×cW×(TA′−20)\Rightarrow 50 \times c \times (90 - {T'_A}) = {m_W} \times {c_W} \times ({T'_A} - 20)---------- (4)
T’A is the initial Temperature of Block B when immersed in water.
Similarly equate (2) and (3) for Block B immersed in water
mBcBΔTB=mWcWΔTW{m_B}{c_B}\Delta {T_B} = {m_W}{c_W}\Delta {T_W}
⇒100×c×(45−TB′)=mW×cW×(TB′−20)\Rightarrow 100 \times c \times (45 - {T'_B}) = {m_W} \times {c_W} \times ({T'_B} - 20)--------- (5)
T’B is the initial Temperature of Block B when immersed in water.
We have one common term in RHS, hence we divide (4) by (5),
⇒50×c×(90−T′A)100×c×(45−T′B)=mW×cW×(T′A−20)mW×cW×(T′B−20)\Rightarrow \dfrac{{50 \times c \times (90 - {{T'}_A})}}{{100 \times c \times (45 - {{T'}_B})}} = \dfrac{{{m_W} \times {c_W} \times ({{T'}_A} - 20)}}{{{m_W} \times {c_W} \times ({{T'}_B} - 20)}}
Cancelling like terms,
⇒(90−T′A)2(45−T′B)=(T′A−20)(T′B−20)\Rightarrow \dfrac{{(90 - {{T'}_A})}}{{2(45 - {{T'}_B})}} = \dfrac{{({{T'}_A} - 20)}}{{({{T'}_B} - 20)}}
Equating Numerator terms,
⇒(90−TA′)=(TA′−20)\Rightarrow (90 - {T'_A}) = ({T'_A} - 20)
⇒110=2TA′\Rightarrow 110 = 2{T'_A}
TA′=55∘C{T'_A} = {55^ \circ }C
Equating denominator terms,
⇒2×(45−TB′)=(TB′−20)\Rightarrow 2 \times (45 - {T'_B}) = ({T'_B} - 20)
⇒110=3TB′\Rightarrow 110 = 3{T'_B}
TB′=36.67∘C{T'_B} = {36.67^ \circ }C

Hence, TA′>TB′{T'_A} > {T'_B}, which means Option(A) is the right answer

Note:
Specific heat of a substance is defined as the amount of heat required to raise the temperature of the unit mass of the system by 1 degree. It is measured in Jg−1K−1{J^{}}{g^{ - 1}}{K^{ - 1}} orcalg−1(∘C)−1ca{l^{}}{g^{ - 1}}{({}^ \circ C)^{ - 1}}. The specific heat of a substance is given by the formula
Q=mcΔTQ = mc\Delta T, where c is the specific heat of the substance.