An object with a mass of $2 kg$ , temperature of $214 ^oC$ , and a specific heat of $12 (KJ)/(kg*K)$ is dropped into a container with $25 L$ 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 · 2018-03-28T08:58:01

The water will not evaporate and the change in temperature is $=39.9^@C$

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 $DeltaT_o=214-T$

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

The specific heat of the object is $C_o=12kJkg^-1K^-1$

The mass of the object is $m_0=2kg$

The volume of water is $V=25L$

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

The mass of the water is $m_w=rhoV=25kg$

$2*12*(214-T)=25*4.186*T$

$214-T=(25*4.186)/(2*12)*T$

$214-T=4.36T$

$5.36=214$

$T=214/5.36=39.9^@C$

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

An object with a mass of 2 kg2 kg, temperature of 214 ^oC214 ^oC, and a specific heat of 12 (KJ)/(kg*K)12 (KJ)/(kg*K) is dropped into a container with 25 L 25 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?