Question

3.2 kg of ice at -10° just melts with m mass of steam at 100°c then m is equal to (s=0.5cal/g°c)

Answer 1

Hope this is right answer

Answer 2

Given that,

Mass of ice = 3.2 kg

Initial temperature =-10°C

Final temperature = 100°C

Specific heat = 0.5 cal/g°C

We need to calculate the value of m

Using formula of heat

$ms\Delta T+mL=Ms\Delta T+ML$

Where, m = mass of ice

M = mass of vapour

s = specific heat capacity

$\Delta T$ = temperature

L = latent heat

Put the value into the formula

$3.2\times10^{3}\times0.5\times(0+10)+3.2\times10^{3}\times80=M(0.5\times(100-0)+540)$

$272000=M\times590$

$M=\dfrac{272000}{590}$

$M=461.0\ g$

$M=0.461\ kg$

Hence, The value of m is 0.461 Kg.