An object with a mass of $4 k g$ $4 kg$ , temperature of $240^{o} C$ $240 ^oC$ , and a specific heat of $5 \frac{J}{k g \cdot K}$ $5 J/(kg*K)$ is dropped into a container with $32 L$ $32 L$ 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 · 2017-06-26T11:38:59

The water does not evaporate and the change in temperature is $=0.04ºC$

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 $DeltaT_o=240-T$

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

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

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

$4*0.005*(240-T)=32*4.186*T$

$240-T=(32*4.186)/(4*0.005)*T$

$240-T=6697.6T$

$6698.6T=240$

$T=240/6698.6=0.04ºC$

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

An object with a mass of 4 kg4kg4 kg, temperature of 240 ^oC240oC240 ^oC, and a specific heat of 5 J/(kg*K)5Jkg⋅K5 J/(kg*K) is dropped into a container with 32 L 32L32 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?