If 980kj of energy as heat are transferred to 6.2l of water at 291 k what will the final temperature be? the specific heat of water is 4.18 j/g•k. assume that 1.0 ml of water equals 1.0 g of water.

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

1. they vary with a higher number because boiling is way hotter than melting

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer