Question

A 155.0 g piece of copper at 168 oc is dropped into 250.0 g of water at 20.9 oc. (the specific heat of copper is 0.385 j/goc.) calculate the final temperature of the mixture. (assume no heat loss to the surroundings.)

Answer 1

The final temperature of the mixture is about 28.8°C

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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!

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Given:

initial temperature of water = t₁ = 20.9°C

specific heat capacity of water = c = 4.186 J/gK

mass of copper = m₁ = 155.0 g

mass of water = m₂ = 250.0 g

initial temperature of copper = t₁' = 168°C

specific heat capacity of copper = c' = 0.385 J/gK

Asked:

final temperature of the mixture = t = ?

Solution:

We can calculate the final temperature of the water by using Conservation of Energy as shown below:

$Q_{lost} = Q_{gain}$

$m_1c'(t_1' - t) = m_2c(t - t_1)$

$155(0.385)(168 - t) = 250(4.186)(t - 20.9)$

$59.675(168 - t) = 1046.5(t - 20.9)$

$10025.4 - 59.675t = 1046.5t - 21871.85$

$t = ( 10025.4 + 21871.85 ) \div ( 59.675 + 1046.5 )$

$t \approx 28.8^oC$

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Learn more

• Efficiency of Engine : brainly.com/question/5597682
• Flow of Heat : brainly.com/question/3010079
• Difference Between Temperature and Heat : brainly.com/question/3821712

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Answer details

Grade: College

Subject: Physics

Chapter: Thermal Physics

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

Answer 2

Note that
The specific heat of water is 4.184 J/(g -C)

Let x = the final equilibrium temperature.

Heat loss by the copper is
(155 g)*(0.385 J/(g -C))*(168 - x C) = 59.675(168 -x) J

Heat gain by the water is
(250 g)*(4.184 J/(g-C))*(x - 20.9 C) = 1046(x - 20.9) J

Because there is no heat loss to the surroundings, therefore
59.675(168 - x) = 1046(x - 20.9)
168 - x = 17.5283(x - 20.9)
18.5283x = 534.3415
x = 28.84 °C

Answer:  28.8 °°C  nearest tenth)