The law of conservation of energy
Answer:
answer is c dhjdjdjdjdjdjdhdhddjdj thats why
Answer:
Final temperature = T₂ = 328.815 K
Explanation:
Given data:
Given energy = 980 KJ = 980×1000= 980000 J
Volume = 6.2 L
Initial temperature =T₁= 291 K
Specific heat of water = 4.18 j /g .K
Final temperature = T₂ = ?
Formula:
Q = m. c . ΔT
ΔT = T₂ - T₁
we will first convert the litter into milliliter
6.2 × 1000 = 6200 mL
It is given in question that
1 mL = 1 g
6200 mL = 6200 g
Now we will put the values in formula,
Q = m. c . (T₂ - T₁)
980000 j = 6200 g . 4.18 j /g .K . (T₂ - 291 K)
980000 j = 25916 j/ k . (T₂ - 291 K)
980000 j / 25916 j/ k = T₂ - 291 K
37.8145 K + 291 K =T₂
T₂ = 328.815 K
the properties of liquids usally wet, wet shows expiation on heating and contracting on cooling
The formula you need is: heat= specific heat x mass x ΔT
specific heat= 0.46 j/g-C
mass= 100.0 grams
ΔT= 40.2 - 15.0= 25.2C
heat= (0.46) x (100.0) x (25.2)= 1159.2 joules or 1200 joules (rounded off)
