An object with a mass of $6 kg$ , temperature of $173 ^oC$ , and a specific heat of $23 J/(kg*K)$ is dropped into a container with $32 L$ 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 · 2018-02-25T13:00:06

The water will not evaporate and the change in temperature is $=0.18^@C$

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 $DeltaT_o=173-T$

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

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

The specific heat of the object is $C_o=0.023kJkg^-1K^-1$

The mass of the object is $m_0=6kg$

The volume of water is $V=32L$

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

The mass of the water is $m_w=rhoV=32kg$

$6*0.023*(173-T)=32*4.186*T$

$173-T=(32*4.186)/(6*0.023)*T$

$173-T=970.7T$

$971.7T=173$

$T=173/971.7=0.18^@C$

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

An object with a mass of 6 kg6 kg, temperature of 173 ^oC173 ^oC, and a specific heat of 23 J/(kg*K)23 J/(kg*K) is dropped into a container with 32 L 32 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?