An object with a mass of $1 k g$ $1 kg$ , temperature of $160^{o} C$ $160 ^oC$ , and a specific heat of $24 \frac{J}{k g \cdot K}$ $24 J/(kg*K)$ is dropped into a container with $27 L$ $27 L$ of water at $0^{o} C$ $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2017-05-13T08:14:29

The water does not evaporate and the change in temperature is $=0.034ºC$

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 $DeltaT_o=160-T$

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

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

$C_o=0.024kJkg^-1K^-1$

$1*0.024*(160-T)=27*4.186*T$

$160-T=(27*4.186)/(1*0.024)*T$

$160-T=4709.25T$

$4710.25T=160$

$T=160/4710.25=0.034ºC$

As $T<100ºC$ , the water does not evaporate.

An object with a mass of 1 kg1kg1 kg, temperature of 160 ^oC160oC160 ^oC, and a specific heat of 24 J/(kg*K)24Jkg⋅K24 J/(kg*K) is dropped into a container with 27 L 27L27 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?