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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Answer details

Grade: College

Subject: Physics

Chapter: Thermal Physics

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

Answer 2

The final temperature of water when a 275 g sample of nickel at 100.0°C is placed in 100.0 mL of water at 22.0°C is 39.6 °C

We'll begin by calculating the mass of the water.

Volume of water = 100 mL

Density of water = 1 g/mL

Mass of water =?

Mass = Density × Volume

Mass of water = 1 × 100

Mass of water = 100 g

Finally, we shall determine the equilibrium (i.e final) temperature of water. This can be obtained as follow:

Mass of Nickel (Mₙ) = 275 g

Temperature of Nickel (Tₙ) = 100 °C

Specific heat capacity of Nickel (Cₙ) = 0.444 J/gK

Mass of water (Mᵥᵥ) = 100 g

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gK

Temperature of water (Tᵥᵥ) = 22 °C

Equilibrium temperature (Tₑ) =?

Heat lost by nickel = Heat gained by water

Qₙ = Qᵥᵥ

MₙCₙ(Tₙ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

275 × 0.444 (100 – Tₑ) = 100 × 4.184 (Tₑ – 22)

122.1 (100 – Tₑ) = 418.4(Tₑ – 22)

Clear bracket

12210 – 122.1Tₑ = 418.4Tₑ – 9204.8

Collect like terms

12210 + 9204.8 = 418.4Tₑ + 122.1Tₑ

21414.8 = 540.5Tₑ

Divide both side by 540.5

Tₑ = 21414.8 / 540.5

Tₑ = 39.6 °C

Thus, the final temperature of water is 39.6 °C

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