Question

the heat must be absorbed by ice of mass 500g at -10℃ to take it to water at 20℃ is (specific heat of ice 2.2 kj/kgK,specific heat of water is 4.2kj/kgK and latent heat of fusion of ice is 300kj/kg)​

Answer 1

Answer:

$Q_t_o_t_a_l=203kJ$

Explanation:

1. Heat gained by ice for 10°C ⇒ 0° C :-

       $Q_1=m(S_i_c_e)(\triangle T)\\\\Q_1=(0.5)(2.2)(10)\\\\\boxed{Q_1=11}----(1)$

2. Heat gained by ice for 0°C(ice) ⇒ 0° C(water) :-

       $Q_2=m(\triangle L_f_u_s_i_o_n)\\\\Q_2=(0.5)(300)\\\\\boxed{Q_2=150}----(2)$

3. Heat gained by water for 0°C⇒20° C :-

       $Q_3=m(S_w_a_t_e_r)(\triangle T)\\\\Q_3=(0.5)(4.2)(20)\\\\\boxed{Q_3=42}----(3)$

Now adding (1), (2) and (3) to get total heat to be absorbed

We get :

$Q_t_o_t_a_l=Q_1+Q_2+Q_3=203kJ$

Hope this answer helped you :)