A copper block of mass 4 kg is heated in a fumance to a temperature text425 texto textC and then placed on a large ice block. The mass of ice that will melt in this process will be (Specific heat of copper =500 J kg-1- oC-1 and heat of fusion of ice =336 kJ kg-1 ):

Question

A copper block of mass 4 kg is heated in a fumance to a temperature text425 texto textC and then placed on a large ice block. The mass of ice that will melt in this process will be (Specific heat of copper =500 J kg-1- oC-1 and heat of fusion of ice =336 kJ kg-1 ): (A) 0.5 kg (B) 1 kg (C) 1.5 kg (D) 2.5 kg

Tardigrade Admin · Accepted Answer

Fall in temperature of copper block when it is placed on the ice block =Δ t=425-0=425 oC Heat lost by copper block when it is placed on the ice block. Q1=m1sΔ T =4× 500× 425=850 kJ Heat gained by ice in melting into m2 kg of water. Q2=m2L =m2× 336 =336 m2 kJ According to calorimetry principle, heat lost = heat gained i.e., 850=336 m2 ∴ m2=(850/336)=2.5 kg

Q. A copper block of mass 4 kg is heated in a fumance to a temperature 425oC and then placed on a large ice block. The mass of ice that will melt in this process will be (Specific heat of copper =500Jkg−1−oC−1 and heat of fusion of ice =336kJkg−1 ):

0.5 kg

1 kg

1.5 kg

2.5 kg

Q. A copper block of mass 4 kg is heated in a fumance to a temperature $425^{o} C$ and then placed on a large ice block. The mass of ice that will melt in this process will be (Specific heat of copper $= 500 J k g^{- 1} -^{o} C^{- 1}$ and heat of fusion of ice $= 336 k J k g^{- 1}$ ):