An object with a mass of $1 kg$ , 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 · 2018-03-31T12:58:04

The water does not evaporate and the change in temperature is $=0.06^@C$

The heat is transferred from the hot object to the cold water.

Let $T=$ final temperature of the object and the water

For the cold water, $Delta T_w=T-0=T$

For the object $DeltaT_o=150-T$

$m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)$

The specific heat of the object is $C_o=0.024kJkg^-1K^-1$

The mass of the object is $m_0=1kg$

The volume of water is $V=15L$

The density of water is $rho=1kgL^-1$

The mass of the water is $m_w=rhoV=15kg$

$1*0.024*(150-T)=15*4.186*T$

$150-T=(15*4.186)/(1*0.024)*T$

$150-T=2616.3T$

$2617.3T=150$

$T=150/2617.3=0.06^@C$

As the final temperature is $T<100^@C$ , the water will not evaporate.

An object with a mass of 1 kg1 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?