An object with a mass of $4 k g$ $4 kg$ , temperature of $261^{o} C$ $261 ^oC$ , and a specific heat of $18 \frac{K J}{k g \cdot K}$ $18 (KJ)/(kg*K)$ is dropped into a container with $39 L$ $39 L$ 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 · 2017-02-23T17:58:54

The water does not evapprate and the final temperature is $= 79.8$ $=79.8$ ºC

Let $T =$ $T=$ final temperature of the object and the water

For the cold water, $Delta T_w=T-0=T$

For the object $DeltaT_o=261-T$

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

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

$C_o=18kJkg^-1K^-1$

$m_0 C_o*(270-T) = m_w* 4.186 *T$

$4*18*(261-T)=39*4.186*T$

$261-T=(39*4.186)/(72)*T$

$261-T=2.27T$

$3.27T=261$

$T=261/3.27=79.8ºC$

The water does not evaporate

An object with a mass of 4 kg4kg4 kg, temperature of 261 ^oC261oC261 ^oC, and a specific heat of 18 (KJ)/(kg*K)18KJkg⋅K18 (KJ)/(kg*K) is dropped into a container with 39 L 39L39 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?