10 kg of ice at -10°C is added to 100 kg of water to lower its temperature from 25°C. Consider no heat exchange to surroundings. The decrement to the temperature of water i $s ________$ °C.
(specific heat of ice = 2100 J/kg·°C, specific heat of water = 4200 J/kg·°C, latent heat of fusion of ice = 3.36 × 10⁵ J/kg)

Question

10 kg of ice at -10°C is added to 100 kg of water to lower its temperature from 25°C. Consider no heat exchange to surroundings. The decrement to the temperature of water i $s ________$ °C.

(specific heat of ice = 2100 J/kg·°C, specific heat of water = 4200 J/kg·°C, latent heat of fusion of ice = 3.36 × 10⁵ J/kg)

G1 Physics · Accepted Answer

Heat required to bring 10 kg ice from -10°C to 0°C is Q₁ = m c(ice) ΔT = 10×2100×10 = 2.10×10⁵ J. Heat required to melt the ice at 0°C is Q₂ = mL = 10×3.36×10⁵ = 3.36×10⁶ J. After melting, the 10 kg water formed from ice warms from 0°C to the final temperature T: Q₃ = 10×4200×T. The original 100 kg water cools from 25°C to T and loses heat Q(lost) = 100×4200×(25-T). With no heat exchange to the surroundings, Q₁+Q₂+Q₃ = Q(lost). Therefore, 2.10×10⁵ +3.36×10⁶ +42000T = 420000(25-T). Solving, T =…