An object with a mass of $3 kg$ , temperature of $360 ^oC$ , and a specific heat of $24 J/(kg*K)$ is dropped into a container with $32 L$ 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 · 2018-03-03T14:08:35

The water does not evaporate and the change in temperature is $=0.19^@C$

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 $DeltaT_o=360-T$

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

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

The specific heat of the object is $C_o=0.024kJkg^-1K^-1$

The mass of the object is $m_0=3kg$

The volume of water is $V=32L$

The density of water is $rho=1kgL^-1$

The mass of the water is $m_w=rhoV=32kg$

$3*0.024*(360-T)=32*4.186*T$

$360-T=(32*4.186)/(3*0.024)*T$

$360-T=1860.4T$

$1861.4T=360$

$T=360/1861.4=0.19^@C$

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

An object with a mass of 3 kg3 kg, temperature of 360 ^oC360 ^oC, and a specific heat of 24 J/(kg*K)24 J/(kg*K) is dropped into a container with 32 L 32 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?