A cold pack containing 200 grams of ice at 0 degrees celcius is placed on an athlete's shoulder. When the pack is removed the temperature of the liquid water is 40 degrees …

Question

asked Jul 16, 2021 231k views

1 Answer

Paul De Barros · Answer 1 · 2021-07-22T18:35:14+0000

Answer:

366.99 kJ

Step-by-step explanation:

Enthalpy of melting of ice = 333.55 kJ

The expression for the calculation of the enthalpy change of a process in which liquid water at 0 °C converts to 40 °C is shown below as:-

$\Delta H=m* C* \Delta T$

Where,
$\Delta H$ is the enthalpy change

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass = 200 g

Specific heat = 4.18 J/g°C

$\Delta T=40-0\ ^0C=40\ ^0C$

So,

$\Delta H=200* 4.18* 40\ J=33440\ J$

Also, 1 J = 0.001 kJ

So,

$\Delta H=33.44\ kJ$

So, total energy absorbed = 333.55 + 33.44 = 366.99 kJ