If 10 g of ice at 0 ∘ C is mixed with 10 g of water at 4 0 ∘ C , the final mass of water in the mixture is

Question

TestHub · Accepted Answer

To find the final mass of water, we first calculate the heat released by water cooling from $4 0^{\circ} C$ to $0^{\circ} C$ :

$Q_{re l e a se d} = m_{w a t er} c_{w a t er} Δ T = (10 g) (1 c a l / g^{\circ} C) (4 0^{\circ} C) = 400 c a l$ .

This heat melts some ice. The heat required to melt ice is $Q_{m e lt} = m_{i ce_m e lt} L_{f}$ .

Given $L_{f} = 80 c a l / g$ , then $400 c a l = m_{i ce_m e lt} (80 c a l / g)$ .

So, $m_{i ce_m e lt} = 5 g$ .

The final mass of water is the initial water plus the melted ice: $10 g + 5 g = 15 g$ .

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