An object with a mass of $2 k g$ $2 kg$ , temperature of $331^{o} C$ $331 ^oC$ , and a specific heat of $18 \frac{K J}{k g \cdot K}$ $18 (KJ)/(kg*K)$ is dropped into a container with $42 L$ $42 L$ 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 · 2017-07-08T05:50:21

The water does not evaporate and the change in temperature is $=56.3ºC$

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 $DeltaT_o=331-T$

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

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

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

$2*18*(331-T)=42*4.186*T$

$331-T=(42*4.186)/(2*18)*T$

$331-T=4.88T$

$5.88T=331$

$T=331/5.88=56.3ºC$

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

An object with a mass of 2 kg2kg2 kg, temperature of 331 ^oC331oC331 ^oC, and a specific heat of 18 (KJ)/(kg*K)18KJkg⋅K18 (KJ)/(kg*K) is dropped into a container with 42 L 42L42 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?