Question

40 g of water at 60°C is poured into a vessel containing 50 g of water at 20°C. The final temperature recorded is 30°C. Calculate the thermal capacity of vessel.

[Take the specific heat capacity of water is 4.2 J g⁻¹°C⁻¹]

Answer 1

Given:

Hot Water:

Mass (m₁) = 40 g

Initial Temperature (T₁) = 60°C

Cold Water and Vessel:

Mass of cold water (m₂) = 50 g

Initial Temperature (T₂) = 20°C

Mixture:

Final Temperature (T_f) = 30°C

Specific heat capacity of water (c_w): = 4.2 J g⁻¹°C⁻¹

Thermal Capacity of the vessel: Let it be C'.

Calculate heat lost by hot water (Q_lost):

Q_lost = m₁ × c_w × (T₁ − T_f)

= 40 × 4.2 × (60 − 30)

= 40 × 4.2 × 30

= 5040 J

Calculate heat gained by cold water (Q_water):

Q_water = m₂ × c_w × (T_f − T₂)

= 50 × 4.2 × (30 − 20)

= 50 × 4.2 × 10

= 2100 J

Express heat gained by the vessel \[(Q_{\text{vessel}})\]:

\(Q_{\text{vessel}}=C^{\prime }\times (T_{f}-T_{2})\)

\(Q_{\text{vessel}}=C^{\prime }\times (30-20)\)

\(Q_{\text{vessel}}=C^{\prime }\times 10\)

3. Apply the Principle of Calorimetry:

\(Q_{\text{lost}}=Q_{\text{water}}+Q_{\text{vessel}}\)

5040 = 2100 + (C' × 10)

Subtract 2100 from both sides:

5040 − 2100 = C' × 10

2940 = C' × 10

\[C^{\prime }=\frac{2940}{10}\]

= 294 J°C⁻¹

The thermal capacity of the vessel is 294 J°C⁻¹.