An object with a mass of $1 kg$ , temperature of $320 ^oC$ , and a specific heat of $14 J/(kg*K)$ 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 · 2018-02-14T09:41:49

The water does not evaporate and the change in temperature is $=0.07^@C$

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 $DeltaT_o=320-T$

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

The specific heat of water is $C_w=4.186kJkg^-1K^-1$

The specific heat of the object is $C_o=0.014kJkg^-1K^-1$

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

The volume of water is $V=16L$

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

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

$1*0.014*(320-T)=16*4.186*T$

$320-T=(16*4.186)/(1*0.014)*T$

$320-T=4784T$

$4785T=320$

$T=320/4785=0.07^@C$

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

An object with a mass of 1 kg1 kg, temperature of 320 ^oC320 ^oC, and a specific heat of 14 J/(kg*K)14 J/(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?