Question

20g of ice and 20g of hot water is mixed, when the ice is melted the temperature of the mixture was found to be $0{}^\circ \text{C}$ . The temperature of hot water taken should be (L of ice = $80\text{ cal/g}$ )
(A) $40{}^\circ C$
(B) $72{}^\circ C$
(C) $80{}^\circ C$
(D) $96{}^\circ \text{C}$ $$

Answer 1

This question is from specific heat capacity and latent heat .Assuming that ice was at $0{}^\circ C$ initially. Then heat lost by the hot water is equal to heat gained by ice. The heat lost by hot water can be calculated from specific heat capacity. Heat gained by the ice can be calculated from latent heat.

Formula used
Specific heat capacity $\text{s}=\dfrac{\vartriangle \theta }{\text{m}\left( \vartriangle \text{T} \right)}$
$\vartriangle \theta =\text{ms}\left( \vartriangle \text{T} \right)$
Heat gained by ice can be obtained from, Latent heat $\text{L}=\dfrac{\theta }{\text{m}}$
$\theta =\text{ mL}$ .

Complete step by step solution
Assuming theta the ice was at $\text{0 }\\!\\!{}^\circ\\!\\!\text{ C}$ initially, ${{T}_{1}}=0{}^\circ C$
Heat lost by the hot water = heat gained by ice.
Since, heat lost by Hot water is calculated by specific heat capacity.
Specific heat capacity $\text{s }=\dfrac{\Delta \theta }{\text{m}\Delta \text{T}}$
m is mass of hot water = 20g
S is specific heat capacity,
For water S = $1\text{cal }{{\text{g}}^{-1}}\text{ }{}^\circ {{\text{C}}^{-1}}$
$\vartriangle \text{T}={{\text{T}}_{2}}-{{\text{T}}_{1}}$
${{\text{T}}_{2}}->$ Temperature of water = T
${{\text{T}}_{1}}->$ Temperature of ice $=0{}^\circ \text{C}$
Latent heat (Heat gained by ice) $\text{L}=\dfrac{\theta }{{{\text{m}}_{2}}}$
${{\text{m}}_{2}}$ is mass of ice =20g
Latent heat of ice $=80\text{cal gh-1s}$
By condition, $\text{m s }(\Delta \text{T})\text{ }=\text{ }{{\text{m}}_{\text{2}}}\text{L}$
Put all the above values $\text{T}=80{}^\circ \text{C}$
$\left( 20 \right)\left( 1 \right)\left( \text{T}-0 \right)~~=20\left( 80 \right)$
Temperature of water is $80{}^\circ \text{C}$ .

Note
Specific heat for Hydrogen is the maximum $=3\cdot 5\text{ cal }{{\text{g}}^{-1}}\text{ }{}^\circ {{\text{C}}^{-1}}$
Specific heat for ice $=0\cdot 5\text{ cal }{{\text{g}}^{-1}}\text{ }{}^\circ {{\text{C}}^{-1}}$
Latent heat of substance is the amount of heat required to change the state of unit mass of the substance from solid to liquid or from liquid to water without any change in temperature.