An object with a mass of $4 kg$ , temperature of $105 ^oC$ , and a specific heat of $2 J/(kg*K)$ is dropped into a container with $16 L$ 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 · 2017-05-19T06:36:21

The water does not evaporate and the change in temperature is $=0.013ºC$

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 $DeltaT_o=105-T$

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

$C_w=4.186kJkg^-1K^-1$

$C_o=0.002kJkg^-1K^-1$

$4*0.002*(105-T)=16*4.186*T$

$105-T=(16*4.186)/(4*0.002)*T$

$105-T=8372T$

$8373T=105$

$T=105/8373=0.013ºC$

As $T<100ºC$ , the water does not evaporate.

An object with a mass of 4 kg4 kg, temperature of 105 ^oC105 ^oC, and a specific heat of 2 J/(kg*K)2 J/(kg*K) is dropped into a container with 16 L 16 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?