Question

The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3 J/mol · K. The conversion of 50.0 g of ice at 0.00°C to liquid water at 22.0°C requires __________ kJ of heat.

Answer 1

Answer:

$21.5$ kJ

Explanation:

$M$ = mass of  1 mole of water = 18 g

$m$ = given mass of water = 50 g

Number of moles of water in the given amount is given as

$n=\frac{m}{M}$

$n=\frac{50}{18}$

$n=2.8$

$L$ = Heat of fusion of water = 6.01 kJ/mol = 6010 J/mol

$c$ = specific heat of water = 75.3 J/mol

$\Delta T$ = Change in temperature = 22.0 °C

Amount of heat required is given as

$Q = nL + n c \Delta T$

$Q = (2.8) (6010) + (2.8) (75.3) (22)$

$Q = 21466.5$ J

$Q = 21.5$ kJ

Answer 2

The quantity of heat required to convert the given ice into liquid water at the given temperature is 21,297.82 J.

The given parameters;

• heat of fusion, ΔH = 6.01 kJ/mol
• heat capacity of liquid water, C = 75.3 J/mol.K
• mass of the water, m = 50 g
• initial temperature, t₁ = 0 ⁰C
• final temperature, t₂ = 22 ⁰C

The quantity of heat required to convert the given ice into liquid water at the given temperature is calculated as;

$Q = \frac{m}{18 _{H_2O}} \Delta H \ + \ \frac{m}{18 _{H_2O}} C\Delta t\\\\Q = \frac{50}{18} \times 6010 \ \ + \ \ \frac{50}{18} \times 75.3 \times (22-0)\\\\ Q= 16,695.78 \ \ + \ 4602.04\\\\ Q = 21,297.82 \ J$

Thus, the quantity of heat required to convert the given ice into liquid water at the given temperature is 21,297.82 J.

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