Question

A block of aluminum at a temperature of T1 = 32 degrees C has a mass of m1 = 12 kg. It is brought into contact with another block of aluminum with a mass of m2 = 0.5m1, at a temperature of T2 = 9 degrees C. This aluminum has a heat capacity of c = 899 J/kg K.

Answer 1

Answer:$T=21.33 ^{\circ}$

Explanation:

Given

mass of First Block $m_1=12 kg$

Temperature $T_1=32^{\circ}$

mass of second block $m_2=0.5 m_1=6 kg$

Temperature $T_2=9^{\circ}$

Heat capacity of aluminium c=899 J/kg-K

Final Temperature acquired by both blocks at steady state

Heat loss first block =Heat gain by second block

$m_1\times c\times (32-T)=m_2\times c\times (T-9)$

$12\times 899\times (32-T)=6\times 899\times (T-9)$

$2\times 32=3T$

$T=\frac{64}{3}=21.33 ^{\circ}$

Answer 2

The final temperature of the aluminum blocks is 24.3° C

What is the final temperature attained by the two aluminum blocks?

The quantity of heat absorbed/evolved is given by the formula:

• Q = mcΔT

where:

• m is mass
• Q = quantity of heat
• c is heat capacity
• ΔT is temperature difference; T - t
• T is final temperature
• t is initial temperature

From the principle; heat gained = heat lost

Heat gained = 0.5 * 12 * 899 * (T - 9)

Heat lost = 12 * 899 * (T - 32)

Equating both sides:

0.5 * 12 * 899 * (T - 9) = 12 * 899 * (32 - T)

5394T - 48546 = 345216 -10788T

10788T + 5394T = 345216 + 48546

5394T = 296670

T = 24.3

Therefore, the final temperature of the aluminum blocks is 24.3° C

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