Question

A copper block of mass 2.5 Kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice blocks. What is the maximum amount of ice that can melt?

Answer 1

$\boxed{\boxed{\sf \red{Values \: to \: be \: noted}}}$

$\rm{Specific \: heat \: of \: copper \: = \: 0.39 \:J {g}^{ - 1} K^{ - 1} }$

$\rm{latent \: heat \: of \: fusion \: of \: water \: = 335\: J {g}^{ - 1} }$

$\boxed{\boxed{\sf \red{Let \: us \: come \: to \: question}}}$

A copper block of mass 2.5 Kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice blocks. What is the maximum amount of ice that can melt ?

$\underline{ \underline{ \sf \red{Answer : }}}$

$\sf{Given}$

$\sf{mass \: of \: copper \: block, m_{1} \: = \: 2.5 \: kg}$

$\sf{Specific \: heat \: of \: copper \: = \: 0.39 \:J {g}^{ - 1} K^{ - 1} = 0.39 J \: {kg}^{ - 1} {C}^{ - 1} }$

$\sf{Temperature \: of \: furnace \: \Delta \theta = 500°C}$

$\sf{latent \: heat \: of \: fusion \: of \: water \: = 335\: J \: {g}^{ - 1} = 335 \times {10}^{3} J \: {kg}^{ - 1} }$

$\sf{If \: Q \: be \: the \: heat \: absorbed \: by \: the \: copper \: block, then}$

$\boxed{ \boxed{ \sf \red{Q = m_{1} S \Delta \theta}}} - - - (1)$

$\sf{Let \: m_{2} \: be \: the \: mass \: of \: ice \: melted \: when \: copper \: block \: is \: placed \: on \: it,}$

$\boxed{\boxed{ \sf \red{ \therefore \: Q = m_{2} L}}} - - - (2)$

$\sf \red{From \: (i) \& (ii), \: we \: get}$

$\sf{m_{1} S \: \Delta \theta = m_{2} L}$

$\sf{ m_{2} = \frac{m_{1} S \: \Delta \theta}{L} }$

$= \frac{2.5 \times 0.39 \times {10}^{3} \times 500 }{335 \times {10}^{3} }$

$= 1.455 \: kg$

$\approx \: 1.5 \: kg$

Answer 2

Answer:

Explanation:

Mass of copper block , m = 2.5 kg = 2500 g

Rise in temperature of the copper block = 500°c

Specific heat of copper , C = 0.39 J/g°C

Heat of fusion of water L =335 J/G

The maximum heat the copper block can lose,

Q=m c

= 2500 x 0.39 x 500

=487500 J

Let m 1 g  be the amount of ice that melts when the copper block is placed on the ice block.  

Q =m  1 L

m 1 Q/L=487500/335=1455.22g

Hence, the maximum amount of ice that can melt is 1.45 kg.

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