An object with a mass of $4 kg$ , temperature of $265 ^oC$ , and a specific heat of $12 (KJ)/(kg*K)$ is dropped into a container with $39 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

Question

1834 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-26T06:34:54

The water will not evaporate and the change in temperature is $=60.2^@C$

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 $DeltaT_o=265-T$

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

The specific heat of water is $C_w=4.186kJkg^-1K^-1$

The specific heat of the object is $C_o=12kJkg^-1K^-1$

The mass of the object is $m_0=4kg$

The mass of the water is $m_w=39kg$

$4*12*(265-T)=39*4.186*T$

$265-T=(39*4.186)/(4*12)*T$

$265-T=3.4T$

$4.4T=265$

$T=265/4.4=60.2^@C$

As the final temperature is $T<100^@C$ , the water will not evaporate

An object with a mass of 4 kg4 kg, temperature of 265 ^oC265 ^oC, and a specific heat of 12 (KJ)/(kg*K)12 (KJ)/(kg*K) is dropped into a container with 39 L 39 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?