This preview shows page 1. Sign up to view the full content.
17.94:
Equating the heat lost be the soda and mug to the heat gained by the ice and
solving for the final temperature
=
T
(
29
(
29
(
29
(
29
(
29 (
29
(
29
(
29 (
29
(
29
(
29
(
29
(
29
(
29
(
29
(
29
K
kg
J
4190
kg
120
.
0
K
kg
J
910
kg
257
.
0
K
kg
J
4190
kg
00
.
2
kg
J
10
334
C
0
.
15
K
kg
J
2100
kg
120
.
0
C
0
.
20
K
kg
J
910
kg
257
.
0
K
kg
J
4190
kg
00
.
2
3
⋅
+
⋅
+
⋅
×
+
°
⋅

°
⋅
+
⋅
C.
1
.
14
°
=
Note that the mass of the ice
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: ( 29 kg 120 . appears in the denominator of this expression multiplied by the heat capacity of water; after the ice melts, the mass of the melted ice must be raised further to . T...
View
Full
Document