Question

Calculate total heat. If mass = 100g specific heat = 0.1

Answer 1

Answer:

1 Answer

1) Net Heat Change in this Process = 0 

2) ice will be converted to water, then all will come at a equilibrium. Temperature. 

Units are in SI system. 

Steps:

1) Given, 

Mass of Caloriemeter, $$m = 100g = 0.1kg$$ 

Specific Heat of Caloriemeter, $$S (C) = 0.1 units$$ 

Mass of liquid, $$m. = 0.25 kg$$ 

Specific Heat of Liquid, $$s(L) = 0.4 units$$ 

Mass of Ice, $$= 0.01kg$$ 

Latent heat of fusion of Ice, $$L(f) = 80 units$$ 

Specific heat of water, $$S(w) = 1 units.$$ 

2) Heat due to fusion $$= m L$$

Heat due to change in $$T = ms* del(T)$$

Let the final Equilibrium Temperature be T. Celsius. 

$$Q(net)=Q(1)+Q(2)+Q(3)+Q(4)0=0.1\ast 0.1(T-30)+0.25\ast 0.4(T-30)+0.01\ast 80+1\ast 0.01(T-0)$$

Multiply by 100 on both sides, 

$$0=T-30+10(T-30)+80+T$$

$$12T=250$$

$$T=20.83°C$$

Therefore, Final Temperature of mixture $$= 20.83°C$$