An object with a mass of $5 kg$ , temperature of $123 ^oC$ , and a specific heat of $14 J/(kg*K)$ is dropped into a container with $32 L$ 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 · 2017-04-30T12:36:29

The water does not evaporate and the change in temperature is $0.05ºC$

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 $DeltaT_o=123-T$

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

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

$C_o=0.014kJkg^-1K^-1$

$5*0.014*(123-T)=32*4.186*T$

$123-T=(32*4.186)/(5*0.014)*T$

$123-T=2511.6T$

$2512.6T=123$

$T=123/2512.6=0.05ºC$

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

An object with a mass of 5 kg5 kg, temperature of 123 ^oC123 ^oC, and a specific heat of 14 J/(kg*K)14 J/(kg*K) is dropped into a container with 32 L 32 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?