An object with a mass of $2 kg$ , temperature of $320 ^oC$ , and a specific heat of $25 J/(kg*K)$ 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?

Question

1489 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-19T06:28:09

The water does not evaporate and the change in temperature is $=0.25ºC$

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 $DeltaT_o=320-T$

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

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

$C_o=0.025kJkg^-1K^-1$

$2*.025*(320-T)=16*4.186*T$

$320-T=(16*4.186)/(2*0.025)*T$

$320-T=1339.5T$

$1340.5T=320$

$T=320/1340.5=0.25ºC$

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

An object with a mass of 2 kg2 kg, temperature of 320 ^oC320 ^oC, and a specific heat of 25 J/(kg*K)25 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?