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]

Given,

• Mass of copper (m_C) = 0.12 kg = 0.12 x 1000 g = 120 g
• Mass of water (m_W) = 0.84 kg = 0.84 x 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 x c_C x (T - T₂)

= 120 x 0.4 x (T - 35)

= 48 x (T - 35)

And

Heat gained by water = m_W x c_W x (T₂ - T₁)

= 840 x 4.2 x (35 - 15)

= 840 x 4.2 x 20

= 70560 J

As,

Heat lost by copper = Heat gained by water

⟹ 48 x (T - 35) = 70560

⟹ T - 35 = $\dfrac{70560}{48}$ = 1470

⟹ T = 1470 + 35 = 1505 °C

Hence, temperature of the flame is 1505 °C.