Problem — Pot on the stove

Problem

An aluminium pot of mass $m_p = 1{,}2 \ \text{kg}$ ($c_p = 900 \ \text{J/(kg·K)}$) contains $1{,}5 \ \text{kg}$ of water at $15 \ ^{\circ}\text{C}$. A hob delivers a power $P = 2 \ \text{kW}$, of which only $60\%$ actually reaches the water (the rest is dispersed). (a) Calculate the time needed to bring the water and pot to $100 \ ^{\circ}\text{C}$. (b) Calculate the subsequent time to vaporise $0{,}3 \ \text{kg}$ of water ($L_v = 2{,}26 \cdot 10^6 \ \text{J/kg}$). (Assume $c_{\text{acqua}} = 4186 \ \text{J/(kg·K)}$.)

Solution

Only $60\%$ of the delivered power is useful, so the power actually available is $P_u = 0{,}60 \cdot 2000 = 1200 \ \text{W}$ (a) The heat needed to warm the pot and water together by $\Delta T = 85 \ \text{K}$ is $Q_a = (m_p c_p + m_w c_w)\,\Delta T = (1{,}2 \cdot 900 + 1{,}5 \cdot 4186)\cdot 85 = (1080 + 6279)\cdot 85 = 7359 \cdot 85 \approx 625500 \ \text{J}$ from which the time $t_a = \frac{Q_a}{P_u} = \frac{625500}{1200} \approx 521 \ \text{s} \approx 8{,}7 \ \text{min}$ (b) The heat needed to vaporise $0{,}3 \ \text{kg}$ of water at constant temperature is $Q_b = m \, L_v = 0{,}3 \cdot 2{,}26 \cdot 10^6 = 678000 \ \text{J}$ from which $t_b = \frac{Q_b}{P_u} = \frac{678000}{1200} = 565 \ \text{s} \approx 9{,}4 \ \text{min}$ $\ev{t_a \approx 8{,}7 \ \text{min} \qquad t_b \approx 9{,}4 \ \text{min}}$

Collegamenti

Argomenti: Termologia Concetti: Calore specifico e capacità termica · Calore latente e cambiamenti di stato