An object with a mass of $12 kg$ , temperature of $145 ^oC$ , and a specific heat of $23 J/(kg*K)$ is dropped into a container with $35 L$ 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 · 2018-01-12T09:12:58

The water does not evaporate and the change in temperature is $=0.27^@C$

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 $DeltaT_o=145-T$

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

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 $m_w=35kg$

$12*0.023*(145-T)=35*4.186*T$

$145-T=(35*4.186)/(12*0.023)*T$

$145-T=530.8T$

$531.8T=145$

$T=145/531.8=0.27^@C$

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

An object with a mass of 12 kg12 kg, temperature of 145 ^oC145 ^oC, and a specific heat of 23 J/(kg*K)23 J/(kg*K) is dropped into a container with 35 L 35 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?