1 kg of water is heated from 40 ° C to 70 ° C, If its volume remains constant, then the change in internal energy is (specific heat of water = 4148 J kg-1K-1)

Question

1 kg of water is heated from 40 ° C to 70 ° C, If its volume remains constant, then the change in internal energy is (specific heat of water = 4148 J kg-1K-1) (A) 2.44× 105 J (B) 1.62× 105 J (C) 1.24× 105 J (D) 2.62× 105 J

Tardigrade Admin · Accepted Answer

Since volume of water remains constant, then work done

Δ W = P d V = 0

According to first pair of thermodynamics

d Q = d U + d W, d U - d Q = m s Δ T

= 1 \times 4148 \times (70 - 40) = 4148 \times 30

= 124440 J

= 1.244 \times 1 0^{5} J

Q. $1$ kg of water is heated from $4 0^{\circ} C$ to $7 0^{\circ} C$ , If its volume remains constant, then the change in internal energy is (specific heat of water $= 4148 J k g^{- 1} K^{- 1})$

$2.44 \times 1 0^{5} J$

$1.62 \times 1 0^{5} J$

$1.24 \times 1 0^{5} J$

$2.62 \times 1 0^{5} J$

Since volume of water remains constant, then work done
$Δ W = P d V = 0$
According to first pair of thermodynamics
$d Q = d U + d W, d U - d Q = m s Δ T$
$= 1 \times 4148 \times (70 - 40) = 4148 \times 30$
$= 124440 J$
$= 1.244 \times 1 0^{5} J$

Q. 1 kg of water is heated from 40∘C to 70∘C, If its volume remains constant, then the change in internal energy is (specific heat of water =4148Jkg−1K−1)

2.44×105J

1.62×105J

1.24×105J

2.62×105J