How much heat is required to change 5 gram ice ( 0° C ) to steam at 100° C ? Latent heat of fusion and vaporization for water are 80 cal/g and 540 cal/g respectively. Specific heat of water is 1 cal/g/k.

Question

How much heat is required to change 5 gram ice ( 0° C ) to steam at 100° C ? Latent heat of fusion and vaporization for water are 80 cal/g and 540 cal/g respectively. Specific heat of water is 1 cal/g/k. (A) 7200 cal (B) 3600 cal (C) 1800 cal (D) 900 cal

Tardigrade Admin · Accepted Answer

underset(. 0 ° C .)I c e oversetq1 arrow underset(. 0 ° C .)W a t e r oversetq2 arrow underset(. 100 ° C .)W a t e r oversetq3 arrow underset(. 100 ° C .)V a p o u r q=q1+q2+q3 =(.80× 5+msdT+540× 5.) cal =(.80× 5+5× 1× 100+540× 5.) cal =3600 cal

Q. How much heat is required to change 5 gram ice ( $0^{\circ} C$ ) to steam at $10 0^{\circ} C$ ? Latent heat of fusion and vaporization for water are 80 cal/g and 540 cal/g respectively. Specific heat of water is 1 cal/g/k.

7200 cal

3600 cal

1800 cal

900 cal

$(0^{\circ} C) I ce \to q_{1} (0^{\circ} C) Wa t er \to q_{2} (10 0^{\circ} C) Wa t er \to q_{3} (10 0^{\circ} C) Va p o u r$
$q = q_{1} + q_{2} + q_{3}$
$= (80 \times 5 + m s d T + 540 \times 5)$ cal
$= (80 \times 5 + 5 \times 1 \times 100 + 540 \times 5)$ cal
$= 3600$ cal

Q. How much heat is required to change 5 gram ice ( 0∘C ) to steam at 100∘C ? Latent heat of fusion and vaporization for water are 80 cal/g and 540 cal/g respectively. Specific heat of water is 1 cal/g/k.