An object with a mass of $24 k g$ $24 kg$ , temperature of $210^{o} C$ $210 ^oC$ , and a specific heat of $7 \frac{J}{k g \cdot K}$ $7 J/(kg*K)$ is dropped into a container with $48 L$ $48 L$ 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 · 2017-03-23T13:47:08

The water does not evaporate and the temperature rise by $0.15ºC$

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 $DeltaT_o=210-T$

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

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

$C_o=0.007kJkg^-1K^-1$

$24*0.007*(210-T)=48*4.186*T$

$210-T=(48*4.186)/(24*0.007)*T$

$210-T=1388.6T$

$1389.6T=210$

$T=210/1389.6=0.15ºC$

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

An object with a mass of 24 kg24kg24 kg, temperature of 210 ^oC210oC210 ^oC, and a specific heat of 7 J/(kg*K)7Jkg⋅K7 J/(kg*K) is dropped into a container with 48 L 48L48 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?