The amount of heat required to convert 1 gm of ice at 0℃ into steam at 100℃ is
Answers
64000 cal of heat is required to convert 1 g of ice into steam at 100°C.
Explanation:
Given data:
Mass of ice "m" = 0.1 k g = 100 g
c w = 1.00 c a l / g
L v = 540 c a l / g
T 1 = 0 ° C
T 2 = 100 °C
Melted ice is liquid water at 0° C . To evaporate water in these conditions it is necessary to heat the water from 0 to 100 degrees Celsius.
Then, evaporate it.
Q = m c w ( T 2 − T 1 ) m L v = m [ c w ( T 2 − T 1 ) + L v ]
= ( 100 g ) [ ( 1.00 c a l / °C g ) ( 100 - 0°C) + 540 cal/ g ]
= 64000 cal
Thus 64000 cal of heat is required to convert 1 g of ice into steam at 100°C
Also learn more
Calculate the amount of heat in calories required to convert 5 grams of ice at zero degree celsius to steam at hundred degree Celsius mixture??
https://brainly.in/question/4007041
Answer:
3.01055 kJ of heat.
Explanation:
GIVEN:
m = 1 g = 0.001 kg
T1 = 0℃ and T2 = 100℃
T2 - T1 = 100 ℃
Data: Lm =334000 (J/kg),
c = 4185.5 J/(kg⋅K), Lv = 2258000 (J/kg).
As The heat required to melt
1g of ice to water can be calculated as
q1 = Lm m
q1 = 334000 (J/kg) 0.001 (kg) = 334 (J) = 334 (J)
The heat required to convert
1g of water at 0℃ to 1g of water at 100 ℃
can be calculated as
q2 = m c (T2 - T1)
q2= 0.001 kg * 4185.5 J/(kg⋅K) * 100 K= 418.55 J
The heat required to convert
1g of water to steam can be calculated as
q3 = Lv m
q3= 2258000 (J/kg) 0.001 (kg) =2258 (J)
So, total amount of heat required = q1 + q2 + q3
334 + 418.55 + 2258 J=
3010.55 J= 3.01055 kJ of heat.