Question

A 0.5 kg block of aluminum (caluminum=900j/kg⋅∘c) is heated to 200∘c. the block is then quickly placed in an insulated tub of cold water at 0∘c (cwater=4186j/kg⋅∘c) and sealed. at equilibrium, the temperature of the water and block are measured to be 20∘c. part a if the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?

Answer 1

To solve this problem, we should recall the law of conservation of energy. That is, the heat lost by the aluminium must be equal to the heat gained by the cold water. This is expressed in change in enthalpies therefore:

- ΔH aluminium = ΔH water

where ΔH = m Cp (T2 – T1)

The negative sign simply means heat is lost. Therefore we calculate for the mass of water (m):

- 0.5 (900) (20 – 200) = m (4186) (20 – 0)

m = 0.9675 kg

 

Using same mass of water and initial temperature, the final temperature T of a 1.0 kg aluminium block is:

- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)

- 900 T + 180,000 = 4050 T

4950 T = 180,000

T = 36.36°C

 

The final temperature of the water and block is 36.36°C

Answer 2

The final temperature of the water and block is 36°C

Further explanation

Specific Heat Capacity is the amount of energy needed to raise temperature of 1 kg body for 1°C.

$\large {\boxed{Q = m \times c \times \Delta t} }$

Q = Energy ( Joule )

m = Mass ( kg )

c = Specific Heat Capacity ( J / kg°C )

Δt = Change In Temperature ( °C )

Let us now tackle the problem!

Given:

mass of aluminium in the first experiment = m₁ = 0.5 kg

specific heat capacity of aluminium = c₁ = 900 J/kg°C

initial temperature of aluminium = t = 200°C

specific heat capacity of water = c₂ = 4186 J/kg°C

final temperature of the first experiment = t₁ = 20°C

mass of aluminium in the second experiment = m₂ = 1.0 kg

Unknown:

final temperature of the second experiment = t₂ = ?

Solution:

First Experiment :

Firstly , we would like to calculate the mass of the water using Conservation of Energy as shown below

$Q_{lost} = Q_{gained}$

$Q_{aluminium} = Q_{water}$

$m_1 \times c_1 \times (t - t_1) = m \times c_2 \times (t_1 - 0)$

$0.5 \times 900 \times (200 - 20) = m \times 4186 \times (20 - 0) )$

$81000 = 83720~m$

$m = \frac{2025}{2093} ~ kg$

Second Experiment :

Using the same formula , we could calculate the final temperature of the water and block in the second experiment

$Q_{lost} = Q_{gained}$

$Q_{aluminium} = Q_{water}$

$m_2 \times c_1 \times (t - t_2) = m \times c_2 \times (t_2 - 0)$

$1.0 \times 900 \times (200 - t_2) = \frac{2025}{2093} \times 4186 \times (t_2 - 0) )$

$1.0 \times 900 \times (200 - t_2) = 4050 ~ t_2$

$180000 - 900 ~ t_2 = 4050 ~ t_2$

$4950 ~ t_2 = 180000$

$t_2 = \frac{180000}{4950}$

$t_2 \approx 36^oC$

Learn more

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Answer details

Grade: College

Subject: Physics

Chapter: Thermal Physics

Keywords: Heat , Temperature , Block , Aluminium , Ice , Cold , Water