Question

An open container holds ice of mass 0.500 kgkg at a temperature of -17.4 âCâC . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 710 J/minuteJ/minute . The specific heat of ice to is 2100 J/kgâKJ/kgâK and the heat of fusion for ice is 334Ã103J/kg3.How much time tmelts passes before the ice starts to melt?From the time when the heating begins, how much time trise does it take before the temperature begins to rise above 0âC?

Answer 1

Answer:

a) t = 235.2minute

b) 260.94minutes

Explanation:

Heat energy is defined as the energy required to change the temperature of a substance by 1kelvin.

It is expressed as H = mc∆t where m is the mass of the water

c is the specific heat capacity of the water/ice

∆t is the change in temperature

Total heat required to melt the ice

H = mLice + mc∆t

Lice is the latent heat of fusion of ice

a) To calculate how much time tmelts passes before the ice starts to melt, we will only calculate heat energy absorb by the ice before it melts.

H = mLice

H = 0.50(334000)

H = 167,000Joules

If heat is supplied to the container at the constant rate of 710 J/minute, the time taken before the ice starts to melt will be:

t = 167,000/710

t = 235.2minutes

b) To calculate how much time trise does it take before the temperature begins to rise above 0°C, we will calculate the total energy absorbed at 0°C first.

H = 0.50(334000) + 0.50(2100)(0-(-17.4)

Note that the ice melts at 0°C which will be the final temperature

H = 167,000+18270

H = 185,270oules

Since heat is supplied to the container at the constant rate of 710 J/minute

710Joules = 1minute

185,270Joules= x

x = 185,270/710

x = 260.94minutes

The time taken before the temperature begins to rise is 260.94minutes:

Answer 2

Answer:

It takes 25.73 minutes before the ice starts to melt, and it takes 235.21 minutes (about 3.9 hours) before the temperature of ice begins to rise above 0°C.

Explanation:

First, we need to know how much heat is required to rise the temperature of ice to 0°C. To do that, we use the following equation (using the equivalence -17.4°C≈255.6°K and 0°C≈273°K):

$Q=mc\Delta T\\\\Q=(0.500kg)(2100J/kg\°K)(273\°K-255.6\°K)\\\\Q=18270J$

So are needed 18,720 Joules of heat. Since there is supplied heat at a constant rate of 710 J/min, we can get the time $t_{melts}$ that passes before the ice startes to melt from the equation:

$v=\frac{Q}{t_{melts}}\\ \\t_{melts}=\frac{Q}{v}\\\\t_{melts}=\frac{18720J}{710J/min}\\\\t_{melts}=25.73min$

Then, it passes 25.73 minutes before the ice starts to melt.

Now, we do the same to get the time it takes before the temperature begins to rise above 0°C $t_{rise}$:

$Q=mL_f\\\\Q=(0.500kg)(334*10^{3}J/kg)\\\\Q=167000J$

Then, are needed 167,000 Joules of heat to completely melt ice. Since a substance in a phase change does not change its temperature, when ice receives this amount of heat then it starts to rise its temperature above 0°C. So, the time $t_{rise}$ is obtained by:

$t_{rise}=\frac{Q}{v}\\\\t_{rise}=\frac{167000J}{710J/min}\\\\t_{rise}=235.21 min$

Finally, it passes 235.21 minutes (about 3.9 hours) before the temperature of ice begins to rise above 0°C.