An object with a mass of $4 kg$ , temperature of $240 ^oC$ , and a specific heat of $5 J/(kg*K)$ is dropped into a container with $15L$ 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 · 2017-09-15T13:45:21

The water does not evaporate and the change in temperature is $=0.08^@C$

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 $DeltaT_o=240-T$

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

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

$C_o=0.005kJkg^-1K^-1$

$m_0=4kg$

$m_w=15kg$

$4*0.005*(240-T)=15*4.186*T$

$240-T=(15*4.186)/(4*0.005)*T$

$240-T=3139.5T$

$3140.5T=240$

$T=240/3140.5=0.08^@C$

As $T<100^@C$ , the water does not evaporate

An object with a mass of 4 kg4 kg, temperature of 240 ^oC240 ^oC, and a specific heat of 5 J/(kg*K)5 J/(kg*K) is dropped into a container with 15L 15L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?