An object with a mass of $2 k g$ $2 kg$ , temperature of $50^{o} C$ $50 ^oC$ , and a specific heat of $19 \frac{J}{k g \cdot K}$ $19 J/(kg*K)$ is dropped into a container with $24 L$ $24 L$ 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 · 2017-02-28T20:43:50

The water does not evaporate. The change in temperature is $=0.16º$ C

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

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

For the object $DeltaT_o=50-T$

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

$C_w=4186Jkg^-1K^-1$

$C_o=19Jkg^-1K^-1$

$m_0 C_o*(50-T) = m_w* 4186 *T$

$2*19*(50-T)=24*4186*T$

$50-T=(24*4186)/(38)*T$

$50-T=306.9T$

$307.9T=50$

$T=50/307.9=0.16ºC$

The water does not evaporate

An object with a mass of 2 kg2kg2 kg, temperature of 50 ^oC50oC50 ^oC, and a specific heat of 19 J/(kg*K)19Jkg⋅K19 J/(kg*K) is dropped into a container with 24 L 24L24 L of water at 0^oC 0oC0^oC . Does the water evaporate? If not, by how much does the water's temperature change?