Assertion: calorimeter's heat | /r/HasProperty | lost to surrounding
Facets:
Source sentences:
text: Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or: [latex]{q}_{\text{rebar}}=-{q}_{\text{water}}[/latex] Since we know how heat is related to other measurable quantities, we have:
text: In this technique, we immerse a steel vessel (commonly called bomb) in a water bath so that no heat is lost to the surrounding.
text: Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or: [latex]q_{\text{rebar}} = -q_{\text{water}}[/latex]
text: Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or: \[q_\ce{rebar}=−q_\ce{water} \label{5.3.3}\]
text: Calorimeters are designed to be well-insulated, so no heat is gained from or lost to the surroundings.
text: Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or: \({q}_{\text{rebar}}=\text{−}{q}_{\text{water}}\)
text: Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = -heat taken in by water, or: [latex]{q}_{\text{rebar}}=\text{-}{q}_{\text{water}}[/latex] Since we know how heat is related to other measurable quantities, we have:
text: Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = − heat taken in by water, or: \[q_\ce{rebar}=−q_\ce{water} \] Since we know how heat is related to other measurable quantities, we have: \[(c×m×ΔT)_\ce{rebar}=−(c×m×ΔT)_\ce{water} \]
text: Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or: [latex]{q}_{\text{rebar}}+{q}_{\text{water}}=0[/latex] [latex]{q}_{\text{rebar}}=-{q}_{\text{water}}[/latex]
text: If 0.750 kg of lead at a temperature of 255degrees Celsius is dropped into the can, what is the final temperature of the system if no heat is lost to the surroundings?
text: Note that the mixture first warms up as the reaction occurs, but will then gradually cool as heat is lost to the surroundings.
text: Assuming that all heat transfer was between the rebar and the water, with no heat “lost” to the surroundings, then heat given off by rebar = −heat taken in by water, or: