An object with a mass of $3 kg$ , temperature of $244 ^oC$ , and a specific heat of $37 (KJ)/(kg*K)$ is dropped into a container with $16 L$ 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 · 2018-06-18T06:36:53

Some water will evaporate.

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 $DeltaT_o=244-T$

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

The specific heat of the object is $C_o=37Jkg^-1K^-1$

The mass of the object is $m_0=3kg$

The volume of water is $V=16L$

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

The mass of the water is $m_w=rhoV=16kg$

$3*37*(244-T)=16*4.186*T$

$244-T=(16*4.186)/(3*37)*T$

$244-T=0.603T$

$1.603T=244$

$T=244/1.603=152.2^@C$

As the final temperature is $T>100^@C$ , some water will evaporate.

An object with a mass of 3 kg3 kg, temperature of 244 ^oC244 ^oC, and a specific heat of 37 (KJ)/(kg*K)37 (KJ)/(kg*K) is dropped into a container with 16 L 16 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?