When M1 gram of ice at -10°C (specific heat = 0.5 cal g-1 ° C-1) is added to M2 gram of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g-1 is:

Question

When M1 gram of ice at -10°C (specific heat = 0.5 cal g-1 ° C-1) is added to M2 gram of water at 50°C, finally no ice is left and the water is at 0°C. The value of latent heat of ice, in cal g-1 is: (A) (5M1/M2) - 50  (B) (50M2/M1)  (C) (50M2/M1) - 5  (D) (5M2/M1) - 5

Tardigrade Admin · Accepted Answer

Heat lost = Heat gain ⇒ M2 × 1 × 50 = M1 × 0.5 × 10 + M1.Lf ⇒ Lf = (50 M2 - 5 M1/M1) = (50 M2 /M1) - 5

Q. When $M_{1}$ gram of ice at $- 1 0^{\circ} C$ (specific heat = $0.5 c a l g^{- 1 \circ} C^{- 1}$ ) is added to $M_{2}$ gram of water at $5 0^{\circ} C$ , finally no ice is left and the water is at $0^{\circ} C$ .
The value of latent heat of ice, in cal $g^{- 1}$ is:

$\frac{5 M _{1}}{M _{2}} - 50$

$\frac{50 M _{2}}{M _{1}}$

$\frac{50 M _{2}}{M _{1}} - 5$

$\frac{5 M _{2}}{M _{1}} - 5$

$He a t l os t = He a t g ain$
$\Rightarrow M_{2} \times 1 \times 50 = M_{1} \times 0.5 \times 10 + M_{1} . L_{f}$
$\Rightarrow L_{f} = \frac{50 M _{2} - 5 M _{1}}{M _{1}}$
$= \frac{50 M _{2}}{M _{1}} - 5$

Q. When M1​ gram of ice at −10∘C (specific heat = 0.5calg−1∘C−1) is added to M2​ gram of water at 50∘C, finally no ice is left and the water is at 0∘C. The value of latent heat of ice, in cal g−1 is:

M2​5M1​​−50

M1​50M2​​

M1​50M2​​−5

M1​5M2​​−5