Question

In an experiment to measure the temperature of the flame of a Bunsen burner, a lump of copper of mass 0.12 kg is heated on the flame for a long time. The copper then is quickly transferred into a beaker of negligible heat capacity containing 0.84 kg of water and the temperature of the water rose from 15°C to 35°C. Calculate the temperature of the flame.

[Given, specific heat capacity of copper = 0.4 J g⁻¹ °C⁻¹, specific heat capacity of water = 4.2 J g⁻¹ °C⁻¹]

Answer 1

Given: Mass of copper (m_C) = 0.12 kg = 0.12 × 1000 g = 120 g

Mass of water (m_W) = 0.84 kg = 0.84 × 1000 = 840 g

Initial temperature of water (T₁) = 15°C

Final temperature of water = Final temperature of copper = (T₂) = 35°C

Specific heat capacity of copper (c_C) = 0.4 J g⁻¹ °C⁻¹

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

Let,

Temperature of the flame = Initial Temperature of copper = T

Now,

Heat lost by copper = m_C × c_C × (T − T₂)

= 120 × 0.4 × (T − 35)

= 48 × (T − 35)

Heat gained by water = m_W × c_W × (T₂ − T₁)

= 840 × 4.2 × (35 − 15)

= 840 × 4.2 × 20

= 70560 J

Heat lost by copper = Heat gained by water

⇒ 48 × (T − 35) = 70560

⇒ T − 35 = `70560/48`​ = 1470

⇒ T = 1470 + 35 = 1505 °C

Hence, the temperature of the flame is 1505 °C.