250 g of water at 30° C is contained in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and it's contents to 5° C. Given specific latent heat of fusion of ice = 336 x 10³ J kg^-1, specific heat capacity of copper = 400 J kg^-1 K^-1, specific heat capacity of water is 4200 J kg^-1 K^-1.

Given,

m_copper = 50 g

m_water = 250 g

Final temperature = 5° C.

Let mass of ice required be m_i.

Heat energy gained by ice at 0° C to convert into water at 0° C
= m_i × L
= m_i x 336 J

Heat energy gained by (m) g of water at 0° C to rise it's temperature to 5° C
= m x c x change in temperature
= m_i x c x (5 - 0) = m_i × 4.2 × 5
= 21 x m_i

Heat energy lost by water at 30° C in cooling to 5° C
= m x c x change in temperature
= 250 × 4.2 × (30 - 5)
= 250 × 4.2 × 25
= 26250 J

Heat energy lost by vessel at 30° C to cool down to 5° C =
= m x c x change in temperature
= 50 × 0.4 × (30 - 5)
= 50 × 0.4 × 25
= 500 J

If there is no loss of heat,

Heat energy gained = Heat energy lost

Substituting the values in the relation above we get,

$(336 \times m$i ) + (21 \times mi) = 26250 + 500 \\[0.5em] 357 \times mi = 26750 \\[0.5em] \Rightarrow mi = \dfrac{26750}{357} \\[0.5em] \Rightarrow m_i = 74.93 g

Hence, required mass of ice = 74.93 g