An object with a mass of $24 kg$ , temperature of $270 ^oC$ , and a specific heat of $8 J/(kg*K)$ is dropped into a container with $48 L$ 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 · 2018-05-30T06:27:34

The water does not evaporate and the change in temperature is $=0.26^@C$

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 $DeltaT_o=270-T$

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

The specific heat of the object is $C_o=0.008Jkg^-1K^-1$

The mass of the object is $m_0=24kg$

The volume of water is $V=48L$

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

The mass of the water is $m_w=rhoV=48kg$

$24*0.008*(270-T)=48*4.186*T$

$270-T=(48*4.186)/(24*0.008)*T$

$270-T=1046.5T$

$1047.5T=270$

$T=270/1047.5=0.26^@C$

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

An object with a mass of 24 kg24 kg, temperature of 270 ^oC270 ^oC, and a specific heat of 8 J/(kg*K)8 J/(kg*K) is dropped into a container with 48 L 48 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?