An object with a mass of $2 kg$ , temperature of $214 ^oC$ , and a specific heat of $13 (KJ)/(kg*K)$ is dropped into a container with $26 L$ 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 · 2017-06-06T06:30:44

The water does not evaporate and the change in temperature is $=0.05ºC$

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 $DeltaT_o=214-T$

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

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

$C_o=0.013kJkg^-1K^-1$

$2*0.013*(214-T)=26*4.186*T$

$214-T=(26*4.186)/(2*0.013)*T$

$214-T=4186T$

$4187T=214$

$T=214/4187=0.05ºC$

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

An object with a mass of 2 kg2 kg, temperature of 214 ^oC214 ^oC, and a specific heat of 13 (KJ)/(kg*K)13 (KJ)/(kg*K) is dropped into a container with 26 L 26 L of water at 0^oC 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?