The correct option is A Ice is 68.75gm, and water is 181.25gm at 0∘C.
The heat, which 0.15kg of water can release when its temperature is changed from 20∘C to 0∘C.
Q1=mWsWΔTW
where, mW is the mass of the water, sW is the specific heat of water and ΔTW is the change in temperature of water.
Given, mW=0.15 kg, sW=1 kcal kg−1.
Q1=0.15×(1×103)×(20−0)
=3000cal ...(1)
Now, heat absorbed by 0.10kg of ice at −10∘C to increase its temperature to 0∘C.
Q2=miSiΔTi
where mi is the mass of the ice, si is the specific heat of ice and ΔTi is the change in temperature of ice.
Given, mi=0.10 kg, si=0.50 kcal kg−1
Q2=0.10×(0.5×103)×[0−(−10)]
=500cal
So, remaining heat,
Q=Q1−Q2=3000−500=2500 cal
Now as latent heat of ice is l=80kcalkg−1, the remaining heat will melt ice only
L=Qm⇒m=Qm=250080=31.25 g of ice.
Initial amount of ice =0.10 kg=100 g
So, the remaining ice,
=100−31.25=68.75g
Initial amount of water =0.15 kg=150 g
Therefore, total water =150+31.25=181.25g
The temperature of the system will be 0∘C.