10 g of ice at $-10^\circ$C is mixed with 10 g of water at $0^\circ$C. The amount of heat required to raise the temperature of mixture to $10^\circ$C is
[amp_mcq option1=”400 cal” option2=”550 cal” option3=”1050 cal” option4=”1200 cal” correct=”option3″]
This question was previously asked in
UPSC NDA-2 – 2019
1. Heat required to raise the temperature of 10 g of ice from -10°C to 0°C:
$Q_1 = m_{ice} \times c_{ice} \times \Delta T_1$
Assuming specific heat of ice $c_{ice} = 0.5 \, \text{cal/g}^\circ\text{C}$.
$Q_1 = 10 \, \text{g} \times 0.5 \, \text{cal/g}^\circ\text{C} \times (0^\circ\text{C} – (-10^\circ\text{C})) = 10 \times 0.5 \times 10 = 50 \, \text{cal}$.
2. Heat required to melt 10 g of ice at 0°C into water at 0°C:
$Q_2 = m_{ice} \times L_{fusion}$
Assuming latent heat of fusion of ice $L_{fusion} = 80 \, \text{cal/g}$.
$Q_2 = 10 \, \text{g} \times 80 \, \text{cal/g} = 800 \, \text{cal}$.
After this step, we have 10 g of water at 0°C.
3. Heat required to raise the temperature of the 10 g of water (from melted ice) from 0°C to 10°C:
$Q_3 = m_{water(ice)} \times c_{water} \times \Delta T_2$
Assuming specific heat of water $c_{water} = 1 \, \text{cal/g}^\circ\text{C}$.
$Q_3 = 10 \, \text{g} \times 1 \, \text{cal/g}^\circ\text{C} \times (10^\circ\text{C} – 0^\circ\text{C}) = 10 \times 1 \times 10 = 100 \, \text{cal}$.
4. Heat required to raise the temperature of the initial 10 g of water from 0°C to 10°C:
$Q_4 = m_{water(initial)} \times c_{water} \times \Delta T_3$
$Q_4 = 10 \, \text{g} \times 1 \, \text{cal/g}^\circ\text{C} \times (10^\circ\text{C} – 0^\circ\text{C}) = 10 \times 1 \times 10 = 100 \, \text{cal}$.
Total heat required = $Q_1 + Q_2 + Q_3 + Q_4 = 50 + 800 + 100 + 100 = 1050 \, \text{cal}$.
– Heat transfer during temperature change involves specific heat ($Q = mc\Delta T$).
– Need to account for heat required for each component (ice and water) and each process (heating ice, melting ice, heating water).