An object with a mass of $1 kg$ , temperature of $350 ^oC$ , and a specific heat of $6 J/(kg*K)$ is dropped into a container with $28 L$ 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 · 2017-11-23T12:29:48

The water does not evaporate and the change in temperature is $=0.018^@C$

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 $DeltaT_o=350-T$

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

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

The specific heat of the object is $C_o=0.006kJkg^-1K^-1$

The mass of the object is $m_0=1kg$

The mass of the water is $m_w=28kg$

$1*0.006*(350-T)=28*4.186*T$

$350-T=(28*4.186)/(1*0.006)*T$

$350-T=19634.7T$

$19635.7T=350$

$T=350/19635.7=0.018^@C$

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

An object with a mass of 1 kg1 kg, temperature of 350 ^oC350 ^oC, and a specific heat of 6 J/(kg*K)6 J/(kg*K) is dropped into a container with 28 L 28 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?