An object with a mass of $1 kg$ , temperature of $170 ^oC$ , and a specific heat of $32 J/(kg*K)$ is dropped into a container with $8 L$ 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 · 2017-03-25T13:39:21

The water does not evaporate and the change in temperature is $0.16ºC$

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 $DeltaT_o=170-T$

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

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

$C_o=0.032kJkg^-1K^-1$

$1*0.032*(170-T)=8*4.186*T$

$170-T=(8*4.186)/(1*0.032)*T$

$170-T=1046.5T$

$1047.5T=170$

$T=170/1047.5=0.16ºC$

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

An object with a mass of 1 kg1 kg, temperature of 170 ^oC170 ^oC, and a specific heat of 32 J/(kg*K)32 J/(kg*K) is dropped into a container with 8 L 8 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?