1
Question
Four cubes of ice at −10∘C, each of one gram, is taken out from a refrigerator and dropped into 150 g of water at 20∘C. Find the temperature of the water when thermal equilibrium is attained. Assume that no heat is lost to the outside and the water equivalent of the container is 46 g. (Specific heat capacity of water =1 cal/g∘C, specific heat capacity of ice =0.5 cal/g∘C, latent heat of fusion of ice =80 cal/g)
Four cubes of ice at −10∘C, each of one gram, is taken out from a refrigerator and dropped into 150 g of water at 20∘C. Find the temperature of the water when thermal equilibrium is attained. Assume that no heat is lost to the outside and the water equivalent of the container is 46 g. (Specific heat capacity of water =1 cal/g∘C, specific heat capacity of ice =0.5 cal/g∘C, latent heat of fusion of ice =80 cal/g)
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Solution
The correct option is B 17.9∘C
Mass of each cube of ice (m1)=1 g
Mass of water (m2)=150 g
Water equivalent of container (mc)=46 g
Initial temperature of water (T1)=20∘C
Initial temperature of ice (T2)=−10∘C
Let T be the temperature of the system at thermal equilibrium.
When the ice cubes are dropped into water, ice gains heat which is released by water.
Total heat gained by each ice cube:
(Q1)= Heat gained by ice for temperature change from −10∘C to 0∘C + Heat gained by ice to just melt + Heat gained by water to reach thermal equilibrium.
=1×0.5×(0−(−10))+1×80+1×1×(T−0)
⇒Q1=85+T
∵ Four ice cubes are dropped into the water, we get
Q′1=4×Q1=340+4T
Now, ice cubes gain the heat lost by water + container system.
Heat lost by water + Heat lost by container:
Q2=m2cw(20−T)+mccw(20−T)
⇒Q2=150×1×(20−T)+46×1(20−T)
⇒Q2=196×(20−T)
From the principle of calorimetry,
Total heat gained by ice = Total heat lost by water + container
i.e Q′1=Q2
⇒340+4T=196×(20−T)
⇒200T=196×20−340⇒T=3580200
⇒T=17.9∘C
Thus, option (b) is the correct answer.
Mass of each cube of ice (m1)=1 g
Mass of water (m2)=150 g
Water equivalent of container (mc)=46 g
Initial temperature of water (T1)=20∘C
Initial temperature of ice (T2)=−10∘C
Let T be the temperature of the system at thermal equilibrium.
When the ice cubes are dropped into water, ice gains heat which is released by water.
Total heat gained by each ice cube:
(Q1)= Heat gained by ice for temperature change from −10∘C to 0∘C + Heat gained by ice to just melt + Heat gained by water to reach thermal equilibrium.
=1×0.5×(0−(−10))+1×80+1×1×(T−0)
⇒Q1=85+T
∵ Four ice cubes are dropped into the water, we get
Q′1=4×Q1=340+4T
Now, ice cubes gain the heat lost by water + container system.
Heat lost by water + Heat lost by container:
Q2=m2cw(20−T)+mccw(20−T)
⇒Q2=150×1×(20−T)+46×1(20−T)
⇒Q2=196×(20−T)
From the principle of calorimetry,
Total heat gained by ice = Total heat lost by water + container
i.e Q′1=Q2
⇒340+4T=196×(20−T)
⇒200T=196×20−340⇒T=3580200
⇒T=17.9∘C
Thus, option (b) is the correct answer.
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