Question

A 300 g glass thermometer initially at 24 ◦C is put into 160 cm3 of hot water at 91 ◦C. Find the final temperature of the thermometer, assuming no heat flows to the surroundings. The specific heat of glass is 0.2 cal/g · ◦ C and of water 1 cal/g · ◦ C.

Answer 1

Answer:

72.72 °C

Explanation:

$m_g$ = Mass of glass = 300 g

$T_g$ = Temperature of glass = 24 °C

$T_w$ = Temperature of water = 91 °C

$m_w$ = Mass of water = $\rho v=1\ g/cm^3\times 160=160\ g$

$T_f$ = Final temperature of the thermometer

$c_g$ = Specific heat of glass = 0.2 cal/g°C

$c_w$ = Specific heat of water = 1 cal/g°C

In the case of a simple calorimetry

$Q_g=m_gc_g(T_f-T_g)$

$Q_w=m_wc_w(T_f-T_w)$

Heat flow is 0

$Q_g+Q_w=0\\\Rightarrow m_gc_g(T_f-T_g)+m_wc_w(T_f-T_w)=0\\\Rightarrow T_f=\frac{m_gc_gT_g+m_wc_wT_w}{m_gc_g+m_wc_w}\\\Rightarrow T_f=\frac{300\times 0.2\times 24+160\times 1\times 91}{300\times 0.2+160\times 1}\\\Rightarrow T_f=72.72^{\circ}$

The final temperature of the thermometer is 72.72 °C

Answer 2

The final temperature of the the glass thermometer initially at 24 °C is put into 160 cm³ of hot water at 91 °C is 72.72 °C

How to determine the final temperature

We can obtain the final temperature of the thermometer by calculating the equilibrium temperature. This can be obtained as follow:

• Mass thermometer (M) = 300 g
• Specific heat capacity of thermometer (C) = 0.2 cal/gºC
• Temperature of thermometer (T) = 24 °C
• Mass of water (Mᵥᵥ) = 160 g
• Temperature of water (Tᵥᵥ) = 91 °C
• Specific heat capacity of the water (Cᵥᵥ) = 1 cal/gºC

• Equilibrium temperature (Tₑ) =?

Heat gained = Heat loss

MC(Tₑ – T) = MᵥᵥC(Tᵥᵥ – Tₑ)

300 × 0.2 (Tₑ – 24) = 160 × 1 (91 – Tₑ)

60(Tₑ – 24) = 160(91 – Tₑ)

Clear bracket

60Tₑ – 1440 = 14560 – 160Tₑ

Collect like terms

60Tₑ + 160Tₑ = 14560 + 1440

220Tₑ = 16000

Divide both side by 220

Tₑ = 16000 / 220

Tₑ = 72.72 °C

Therefore, the final temperature of the thermometer is 72.72 °C

Learn more about heat transfer:

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