Question

A 275-g sample of nickel at 100.0°C is placed in 100.0 mL of water at 22.0°C. What is the final temperature of the water? Assume that no heat is lost to or gained from the surroundings. Specific heat capacity of nickel = 0.444 J/(g·K)

Answer 1

The final temperature of the water is about 39.6°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 = 22.0°C

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

volume of water = V = 100.0 mL

mass of nickel = m₁ = 275 g

initial temperature of nickel = t₁' = 100.0°C

specific heat capacity of nickel = c' = 0.444 J/gK

Asked:

final temperature of the water = t = ?

Solution:

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

$Q_{lost} = Q_{gain}$

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

$m_1c'(t_1' - t) = \rho V c(t - t_1)$

$275(0.444)(100 - t) = 1 (100) (4.186)(t - 22)$

$122.1(100 - t) = 418.6(t - 22)$

$12210 - 122.1t = 418.6t - 9209.2$

$12210 + 9209.2 = 122.1t + 418.6t$

$540.7t = 21419.2$

$t \approx 39.6^oC$

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Learn more

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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

Answer:

39.6138 °C

Explanation:

Heat gain by water = Heat lost by nickel

Thus,  

$m_{water}\times C_{water}\times (T_f-T_i)=-m_{nickel}\times C_{nickel}\times (T_f-T_i)$

Where, negative sign signifies heat loss

Or,  

$m_{water}\times C_{water}\times (T_f-T_i)=m_{nickel}\times C_{nickel}\times (T_i-T_f)$

For water:

Volume = 100.0 mL

Density of water= 1 g/mL

So, mass of the water:  

$Mass\ of\ water=Density \times {Volume\ of\ water}$  

$Mass\ of\ water=1 g/mL \times {100.0\ mL}$  

Mass of water  = 100 g

Initial temperature = 22.0 °C

Specific heat of water = 4.186 J/g°C

For nickel:

Mass = 275 g

Initial temperature = 100 °C

Specific heat of nickel = 0.444 J/gK = 0.444 J/g°C

So,  

$100\times 4.186\times (T_f-22.0)=275\times 0.444\times (100-T_f)$

$418.6\times (T_f-22.0)=122.1\times (100-T_f)$

$418.6\times T_f+122.1\times T_f=21419.2$

$T_f=39.6138\ ^{0}C$

Thus,  

The final temperature of the combined metals is 39.6138 °C