Question

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.

Answer 1

Final temperature = T₂ = 328.815 K

Step-by-step 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