An open container holds ice of mass 0.550 kg at a temperature of -15.3 âC . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 900 J/minute .The specific heat of ice to is 2100 J/kgâK and the heat of fusion for ice is 334Ã103J/kg.a) How much time tmelts passes before the ice starts to melt?tmelts=___minutesb)From the time when the heating begins, how much time trise does it take before the temperature begins to rise above 0âC?trise=____minutes

Respuesta :

Answer:

a) 201.11minutes

b) 223.75minutes

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 = 0.55(334000)

H = 183,700Joules

If heat is supplied to the container at the constant rate of 900 J/minute, the time taken before the ice starts to melt will be:

t = 183700/900

t = 201.11minutes

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 = 0.55(334000) + 0.55(2100)(0-(-15.3)

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

H = 183700+17671.5

H = 201,371.5Joules

Since heat is supplied to the container at the constant rate of 900 J/minute

900Joules = 1minute

201,371.5Joules= x

900x = 201,371.5

x = 201,371.5/900

x =223.75minutes

The time taken before the temperature begins to rise is 223.75minutes

A) The time that passes before the ice starts to melt is; t_melts = 204.11 minutes

B) The amount of time before the temperature begins to rise above 0°C is; t_rise = 223.75 minutes

We are given;

mass; m = 0.55 kg

Initial temperature; T₁ = -15.3 °C

Final temperature; T₂ = 0 °C

Specific heat capacity of ice; c = 2100 J/kg.K

Specific latent heat of fusion; L = 334000 J/kg.

Rate at which heat is supplied; dq/dt = 900 J/min

A) Let us first find the heat absorbed by the ice before it melts from the formula;

Q = mL

Q = 0.55 × 334000

Q = 183,700 J

Thus, time that passes before the ice starts to melt is;

t = Q/(dq/dt)

t = 183700/900

t_melts = 204.11 minutes

B) To find how much time the time rises before the temperature begins to rise above 0°C,let us first calculate the total heat required to melt the ice;

H = mL + mc∆t

H = 183700 + (1.67 × 2100 × (0 - (-15.3))

H = 201371.5 J

Thus;

t_rise = H/(dq/dt)

t_rise =  201371.5/900

t_rise = 223.75 minutes

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