A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500° C and then placed on large ice block. The maximum amount (in kg ) of ice that can melt is (Specific heat of copper =0.39 Jg -1 °C -1, latent heat of fusion of water =335 Jg -1 )

Question

A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500° C and then placed on large ice block. The maximum amount (in kg ) of ice that can melt is (Specific heat of copper =0.39 Jg -1 °C -1, latent heat of fusion of water =335 Jg -1 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

mIL f= m c S c Δ T c ⇒ m I =( m c S C Δ T C / L f ) =(2.5 × 0.39 ×(500-0)/335) =1.455 Kg

Q. A copper block of mass $2.5 k g$ is heated in a furnace to a temperature of $50 0^{\circ} C$ and then placed on large ice block. The maximum amount (in $k g$ ) of ice that can melt is (Specific heat of copper $= 0.39 J g^{- 1}^{\circ} C^{- 1}$ , latent heat of fusion of water $= 335 J g^{- 1}$ )

$m_{I} L_{f} = m_{c} S_{c} Δ T_{c}$
$\Rightarrow m_{I} = \frac{m _{c} S _{C} Δ T _{C}}{L _{f}}$
$= \frac{2.5 \times 0.39 \times ( 500 - 0 )}{335}$
$= 1.455 K g$

Q. A copper block of mass 2.5kg is heated in a furnace to a temperature of 500∘C and then placed on large ice block. The maximum amount (in kg ) of ice that can melt is (Specific heat of copper =0.39Jg−1∘C−1, latent heat of fusion of water =335Jg−1 )

mI​Lf​=mc​Sc​ΔTc​ ⇒mI​=Lf​mc​SC​ΔTC​​ =3352.5×0.39×(500−0)​ =1.455Kg

Q. A copper block of mass $2.5 k g$ is heated in a furnace to a temperature of $50 0^{\circ} C$ and then placed on large ice block. The maximum amount (in $k g$ ) of ice that can melt is (Specific heat of copper $= 0.39 J g^{- 1}^{\circ} C^{- 1}$ , latent heat of fusion of water $= 335 J g^{- 1}$ )

$m_{I} L_{f} = m_{c} S_{c} Δ T_{c}$
$\Rightarrow m_{I} = \frac{m _{c} S _{C} Δ T _{C}}{L _{f}}$
$= \frac{2.5 \times 0.39 \times ( 500 - 0 )}{335}$
$= 1.455 K g$