5 g of ice at \(-30^\circ C\) and 20 g of water at \(35^\circ C\) are mixed together in a calorimeter. The final temperature of the mixture is
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In calorimetry, use heat lost by hot body = heat gained by cold body. For ice below \(0^\circ C\), first heat it to \(0^\circ C\), then melt it, and then warm the melted water if heat remains.
Step 1: Calculate heat needed to warm 5 g of ice from $-30^\circC$ to $0^\circC$. Using $Q = mc\Delta T$ with specific heat of ice $c_{\text{ice}} = 0.5$ cal/g^\circC: \[ Q_1 = 5 \times 0.5 \times 30 = 75 \text{ cal} \] Step 2: Calculate heat needed to melt the ice at $0^\circC$. Using latent heat of fusion $L = 80$ cal/g: \[ Q_2 = 5 \times 80 = 400 \text{ cal} \] Total heat required by ice: $Q_1 + Q_2 = 75 + 400 = 475$ cal. Step 3: Calculate heat released by 20 g of water cooling from $35^\circC$ to $0^\circC$. Using specific heat of water $c_w = 1$ cal/g^\circC: \[ Q_3 = 20 \times 1 \times 35 = 700 \text{ cal} \] Step 4: Determine whether all ice melts. Heat available from water ($700$ cal) $>$ heat needed to melt all ice ($475$ cal). So all the ice melts. Remaining heat = $700 - 475 = 225$ cal. This remaining heat will raise the temperature of the total water (now 25 g). Step 5: Find the final temperature. Let final temperature be $T^\circC$ above $0^\circC$. The remaining 225 cal warms 25 g of water: \[ 225 = 25 \times 1 \times T \] \[ T = \frac{225}{25} = 9^\circC \] Step 6: State the final answer. The final temperature of the mixture is $9^\circC$. \[ \boxed{9^\circC} \]
