5 ~g of ice at -30^ C and 20 ~g of water at 35^ C are mixed together in a calorimeter. The final temperature of the mixture is (Neglect heat capacity of the calorimeter, specific heat capacity of ice =0.5 cal g ⁻¹ ^ C ⁻¹ and latent heat of fusion of ice =80 cal g ⁻¹ and specific hea: capacity of water .=1 cal g ⁻¹ ^ C ⁻¹ )

Step-by-step solution

Given, s_w=1 cal g ⁻¹ ^ C ⁻¹ aligned & s_ ice =0.5 cal g ⁻¹ ^ C ⁻¹ & L_f=80 cal g ⁻¹ aligned Mass of ice, m_ ice =5 ~g Mass of water, m_w=20 ~g Let final temperature of the mixture be T^ C . According to principle of calorimetry, Heat lost by water = Heat gained by ice aligned m_w s_w(35-T)=m_ ice s_ ice [0- & (-30)] & + m _ ice L_f+m_ ice s_w T aligned 20 1 (35-T)=5 0.5 30+5 80+5 0.5 T On solving, we get T=9^ C