An object with a mass of $3 kg$ , temperature of $155 ^oC$ , 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 · 2017-02-19T06:46:36

The final temperature of the water is $=99$ ºC

The heat transferred from the hot object, is equal to the heat absorbed by 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 $DeltaT_o=155-T$

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

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

$C_o=37kJkg^-1K^-1$

$m_0 C_o*(155-T) = m_w* 4.186 *T$

$3*37*(155-T)=15*4.186*T$

$155-T=(15*4.186)/(3*37)*T$

$155-T=0.566T$

$1.566T=155$

$T=155/1.566=99ºC$

An object with a mass of 3 kg3 kg, temperature of 155 ^oC155 ^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?