An object with a mass of $2 kg$ , temperature of $150 ^oC$ , and a specific heat of $24 J/(kg*K)$ is dropped into a container with $15 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2017-08-19T07:53:57

The water doesn't evaporate

$V = 15L=0.015m^(-3)$
$rho = 1000kg*m^(-3)$

$rho=m/V$

$m = 0.015*1000$
$rArr m = 15kg$

Let final temperature of object and water be $T_f$

According to the principle of calorimetry,

$Q_o = Q_w$
$2*24*(150-T_f) = 15*4.2*T_f$
$rArr T_f ~~64.8^o C$

$:.$ The water wouldn't evaporate and $DeltaT ~~ 65^oC$

An object with a mass of 2 kg2 kg, temperature of 150 ^oC150 ^oC, and a specific heat of 24 J/(kg*K)24 J/(kg*K) is dropped into a container with 15 L 15 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?