An object with a mass of $3 kg$ , temperature of $185 ^oC$ , and a specific heat of $37 (KJ)/(kg*K)$ is dropped into a container with $15 L$ of water at $0^oC$ . Does the water evaporate? If not, by how much does the water's temperature change?

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Answer 1 · 2017-04-13T06:14:39

The water does not evaporate and the change in temperature is $=0.33ºC$

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 $DeltaT_o=185-T$

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

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

$C_o=0.037kJkg^-1K^-1$

$3*0.037*(185-T)=15*4.186*T$

$185-T=(15*4.186)/(3*0.037)*T$

$185-T=565.7T$

$566.7T=185$

$T=185/566.7=0.33ºC$

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

An object with a mass of 3 kg3 kg, temperature of 185 ^oC185 ^oC, and a specific heat of 37 (KJ)/(kg*K)37 (KJ)/(kg*K) is dropped into a container with 15 L 15 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?