Question

please please answer it​

Answer 1

Answer:-

$ΔQ ≈ 725 \: cal ≈ 3045 J$

Explanation:-

Given:-

• $m = 1g$
• $T_1 = -10 \degree C$
• $T_2 = 100 \degree C$
• $s_{ice} = \dfrac{1}{2}\dfrac{cal}{gm\degree C}$
• $s_{water} = 1\dfrac{cal}{gm\degree C}$
• $L_{ice} = 80 \dfrac{cal}{gm}$
• $L_{steam} = 540\dfrac{cal}{gm}$

To Find:

Amount of heat required

Solution:-

w.k.t

ΔQ = msΔT

To convert ice at -10°c to 0°c

$ΔQ_1 = 1g \times \dfrac{1}{2}\dfrac{cal}{gm \degree C} \times (0-(-10)\degree C$

$ΔQ_1 = 1\cancel{g} \times \dfrac{1}{2}\dfrac{cal}{\cancel{gm \degree C}} \times [10]\cancel{\degree C}$

$ΔQ_1 = \dfrac{1}{2} \times 10 \: cal$

$ΔQ_1 = 5 \: cal$

To Convert ice into water at 0°c

ΔQ = mL

$ΔQ_2= 1gm \times 80 \dfrac{cal}{gm }$

$ΔQ_2 = 1\cancel{gm} \times 80 \dfrac{cal}{\cancel{gm }}$

$ΔQ_2 =80 \: cal$

To convert Water at 0 to 100°C

$ΔQ_3 = 1g \times 1 dfrac{cal}{gm \degree C} \times (100-0)\degree C$

$ΔQ_3 = 1\cancel{g} \times 1 \dfrac{cal}{\cancel{gm \degree C}} \times [100]\cancel{\degree C}$

$ΔQ_3 =100 \: cal$

To convert water to steam at 100°C

$ΔQ_4 = 1gm \times 540\dfrac{cal}{gm }$

$ΔQ_4 = 1\cancel{gm} \times 540 \dfrac{cal}{\cancel{gm }}$

$ΔQ_4 =540 \: cal$

Total heat

$ΔQ = ΔQ_1 + ΔQ_2 + Δ Q_3 + ΔQ_4$

$ΔQ =( 5 + 80+ 100+ 540) cal$

$ΔQ = 725cal$

$ΔQ = 725 \times 4.2 J$

$ΔQ ≈ 3045 J$

Answer 2

Explanation:

see pic for the ans. thank u