An object with a mass of $12 k g$ $12 kg$ , temperature of $175^{o} C$ $175 ^oC$ , and a specific heat of $23 \frac{J}{k g \cdot K}$ $23 J/(kg*K)$ is dropped into a container with $35 L$ $35 L$ 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 · 2018-01-17T06:31:21

The water will not evaporate and the change in temperature is $= {0.33}^{\circ} C$ $=0.33^@C$

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 $DeltaT_o=175-T$

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

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 $m_0=12kg$

The mass of the water is $m_w=35kg$

$12*0.023*(175-T)=35*4.186*T$

$175-T=(35*4.186)/(12*0.023)*T$

$175-T=530.8T$

$531.8T=175$

$T=175/531.8=0.33^@C$

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

An object with a mass of 12 kg12kg12 kg, temperature of 175 ^oC175oC175 ^oC, and a specific heat of 23 J/(kg*K)23Jkg⋅K23 J/(kg*K) is dropped into a container with 35 L 35L35 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?