Question

pls answer this question​

Answer 1

The 3 kg of ice at -12°C has to be converted to water at 100°C by the following steps:

1. Converting 3 kg of ice at -12°C to that at 0°C

Here,

$\longrightarrow\ \sf{m=3\ kg}\\\\\\\longrightarrow\ \sf{c_i=2100\ J\ kg^{-1}\ K^{-1}}\\\\\\\longrightarrow\ \sf{\Delta T_1=0-(-12)=12\ K}$

[Note : Temperature "difference" in °C and K are same.]

Then the heat required for this conversion,

$\longrightarrow\ \sf{q_1=mc_i\Delta T_1}\\\\\\\longrightarrow\ \sf{q_1=3\times 2100\times 12}\\\\\\\longrightarrow\ \sf{q_1=75600\ J}$

2. Conversion of ice at 0°C to water at 0°C

[Note : Melting is done at 0°C since the melting point of water is 0°C]

Here total heat required for fusion is to be calculated.

Thus the heat required for this conversion,

$\longrightarrow\ \sf{q_2=m\cdot L_f}\\\\\\\longrightarrow\ \sf{q_2=3\times 3.35\times10^5}\\\\\\\longrightarrow\ \sf{q_2=1005000\ J}$

3. Conversion of water at 0°C to that at 100°C

Here,

$\longrightarrow\ \sf{m=3\ kg}\\\\\\\longrightarrow\ \sf{c_w=4186\ J\ kg^{-1}\ K^{-1}}\\\\\\\longrightarrow\ \sf{\Delta T_2=100-0=100\ K}$

[Note : Temperature "difference" in °C and K are same.]

Then the heat required for this conversion,

$\longrightarrow\ \sf{q_3=mc_w\Delta T_2}\\\\\\\longrightarrow\ \sf{q_3=3\times 4186\times 100}\\\\\\\longrightarrow\ \sf{q_3=1255800\ J}$

Now, the total amount of heat required for the conversion is,

$\longrightarrow\ \sf{q=q_1+q_2+q_3}\\\\\\\longrightarrow\ \sf{q_3=75600+1005000+1255800}\\\\\\\longrightarrow\ \sf{\underline {\underline {q=2336400\ J}}}\\\\\\\longrightarrow\ \sf{\underline {\underline {q=2336.4\ kJ}}}$