An object with a mass of $1 kg$ , temperature of $350 ^oC$ , and a specific heat of $14 J/(kg*K)$ is dropped into a container with $28 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

Question

1489 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-10T06:16:27

The water does not evaporate and the change in temperature is $=0.042ºC$

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 $DeltaT_o=350-T$

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

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

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

$1*0.014*(350-T)=28*4.186*T$

$200-T=(28*4.186)/(1*0.014)*T$

$350-T=8372T$

$8373T=350$

$T=350/8373=0.042ºC$

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

An object with a mass of 1 kg1 kg, temperature of 350 ^oC350 ^oC, and a specific heat of 14 J/(kg*K)14 J/(kg*K) is dropped into a container with 28 L 28 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?