An object with a mass of $3 k g$ $3 kg$ , temperature of $341^{o} C$ $341 ^oC$ , and a specific heat of $12 \frac{K J}{k g \cdot K}$ $12 (KJ)/(kg*K)$ is dropped into a container with $42 L$ $42 L$ of water at $0^{o} C$ $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2018-04-19T13:16:21

The water will not evaporate and the change in temperature is $= {58.0}^{\circ} C$ $=58.0^@C$

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

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

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

For the object $DeltaT_o=341-T$

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

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

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

The volume of water is $V=42L$

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

The mass of the water is $m_w=rhoV=42kg$

$3*12*(341-T)=42*4.186*T$

$341-T=(42*4.186)/(3*12)*T$

$341-T=4.88T$

$5.88T=341$

$T=341/5.88=58.0^@C$

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

An object with a mass of 3 kg3kg3 kg, temperature of 341 ^oC341oC341 ^oC, and a specific heat of 12 (KJ)/(kg*K)12KJkg⋅K12 (KJ)/(kg*K) is dropped into a container with 42 L 42L42 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?