Question

1. A glass weighing 30 g contains 85 g of water at 25⁰C. How much ice at 0⁰C will be needed to bring the temperature down to 5⁰C? The specific heat of glass is 0.16 cal/g, and that if water is 1.00 cal/g.

Answer 1

The final temperature is t.

Heat given to the ice by the glass and water:

$$Q_1=c_gm_g(t-t_g)+c_wm_w(t-t_w).$$

Heat 'received' by the ice:

$$Q_2=m\lambda+c_im(t-t_i).$$

We know that

$$Q_1+Q_2=0,\\ c_gm_g(t-t_g)+c_wm_w(t-t_w)+m\lambda+c_im(t-t_i)=0,\\\space\\ m=\frac{c_gm_g(t-t_g)+c_wm_w(t-t_w)}{c_i(t_i-t)-\lambda}=22\text{ g}.$$

Note: latent heat of fusion of ice is 80 cal/g, specific heat of ice is 0.5 cal/(gºC).